# Exercises: Convergence criteria for series – Serlo

Zur Navigation springen Zur Suche springen

## Application of convergence criteria

Exercise (Convergence proof training 1)

Investigate, whether the following series are divergent, convergent or even absolutely convergent.

1. ${\displaystyle \sum _{k=1}^{\infty }{\frac {(k^{k})^{2}}{k^{(k^{2})}}}}$
2. ${\displaystyle \sum _{k=1}^{\infty }{\frac {(k!)^{2}}{(2k)!}}}$
3. ${\displaystyle \sum _{k=1}^{\infty }{\frac {k}{1+k^{2}}}}$
4. ${\displaystyle \sum _{k=1}^{\infty }\left({\frac {k}{k+1}}\right)^{k}}$
5. ${\displaystyle \sum _{k=1}^{\infty }\left({\frac {k}{k+1}}\right)^{k^{2}}}$
6. ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k}({\sqrt {k+1}}-{\sqrt {k}})}$
7. ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k}}{\sqrt[{k}]{k}}}}$
8. ${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k}(1+{\tfrac {1}{k}})^{k}}{k}}}$

Solution (Convergence proof training 1)

1. Root test: The ${\displaystyle k}$ in the exponents is tedious. We can remove it from one exponent by taking the ${\displaystyle k}$-th root:

{\displaystyle {\begin{aligned}{\sqrt[{k}]{|a_{k}|}}&={\sqrt[{k}]{\left|{\frac {(k^{k})^{2}}{k^{(k^{2})}}}\right|}}={\frac {\sqrt[{k}]{k^{2k}}}{\sqrt[{k}]{k^{(k^{2})}}}}={\frac {k^{2}}{k^{k}}}={\frac {1}{k^{k-2}}}\longrightarrow 0<1\end{aligned}}}

The root test now tells us that the series converges absolutely. Intuitively, the exponent ${\displaystyle (\cdot )^{k^{2}}}$ produces a much faster growth than ${\displaystyle (\cdot )^{2k}}$ . So fast that ${\displaystyle |a_{k}|}$ decays faster than a geometric series ${\displaystyle b_{k}=q^{k},\quad 0\leq q<1}$.

2. Ratio test: By which factor do the sequence elements ${\displaystyle |a_{k}|}$ increase, if we go from ${\displaystyle |a_{k}|}$ to ${\displaystyle |a_{k+1}|}$? If that factor is lower than some constant ${\displaystyle q<1}$, we have absolute convergence:

{\displaystyle {\begin{aligned}\left|{\frac {a_{k+1}}{a_{k}}}\right|&={\frac {\frac {((k+1)!)^{2}}{(2k+2)!}}{\frac {(k!)^{2}}{(2k)!}}}={\frac {((k+1)!)^{2}(2k)!}{(k!)^{2}(2k+2)!}}={\frac {k!k!(k+1)^{2}(2k)!}{k!k!(2k)!(2k+1)(2k+2)}}={\frac {k^{2}+2k+1}{4k^{2}+6k+2}}\\&={\frac {1+{\frac {2}{k}}+{\frac {1}{k^{2}}}}{4+{\frac {6}{k}}+{\frac {2}{k^{2}}}}}\longrightarrow {\frac {1}{4}}<1\end{aligned}}}

So the increase factor is smaller than ${\displaystyle q=1/4}$ and the series converges absolutely.

3. Diract comparison test: For large ${\displaystyle k}$, the series elements essentially behave like ${\displaystyle a_{k}\approx {\frac {k}{k^{2}}}={\frac {1}{k}}}$, which is a divergent harmonic series. So we suspect that this series also diverges. In fact, the ${\displaystyle +1}$ in the denominator makes ${\displaystyle a_{k}}$ smaller than ${\displaystyle {\frac {1}{k}}}$. But for ${\displaystyle k\to \infty }$, ${\displaystyle a_{k}}$ will approach it arbitrarily close and get larger that ${\displaystyle c\cdot {\frac {1}{k}}}$ for any ${\displaystyle 0\leq c<1}$. For convenience, we choose ${\displaystyle c=1/2}$ and get that eventually:

• ${\displaystyle |a_{k}|={\frac {k}{1+k^{2}}}\geq {\frac {k}{k^{2}+k^{2}}}={\frac {k}{2k^{2}}}={\frac {1}{2k}}}$
• But the harmonic series ${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {1}{2k}}}$ diverges.

The series ${\displaystyle a_{k}}$ is "even larger" and therefore also diverges.

4. Term test: What happens for ${\displaystyle k\to \infty }$? Actually, the series elements get close to a constant:

${\displaystyle a_{k}=\left({\frac {k}{k+1}}\right)^{k}={\frac {1}{\left({\frac {k+1}{k}}\right)^{k}}}={\frac {1}{\left(1+{\frac {1}{k}}\right)^{k}}}\to {\frac {1}{e}}\neq 0.}$

The series would intuitively evaluate to ${\displaystyle \sum _{k}a_{k}\approx {\frac {1}{e}}\infty =\infty }$ and is in mathematical words divergent by the term test.

5. Root test: For large ${\displaystyle k}$, we have seen in 4., that ${\displaystyle \left({\frac {k}{k+1}}\right)^{k}}$ goes like ${\displaystyle {\frac {1}{e}}}$. So ${\displaystyle \left({\frac {k}{k+1}}\right)^{k^{2}}}$ should behave like ${\displaystyle {\frac {1}{e^{k}}}}$, which is a geometric series and hence convergent. We can verify this convergence by the root criterion:

${\displaystyle {\sqrt[{k}]{|a_{k}|}}=\left({\frac {k}{k+1}}\right)^{k}={\frac {1}{\left({\frac {k+1}{k}}\right)^{k}}}={\frac {1}{\left(1+{\frac {1}{k}}\right)^{k}}}\to {\frac {1}{e}}<1.}$

Hence, the series converges absolutely.

6. Alternating series test:

The ${\displaystyle (-1)^{k}}$ suggests that we have to deal with an alternating series. And since ${\displaystyle {\sqrt {k+1}}-{\sqrt {k}}}$ is always positive, this is indeed true. Further, the two square roots are moving very close together for high ${\displaystyle k}$. For instance ${\displaystyle {\sqrt {1}}-{\sqrt {0}}=1}$, but ${\displaystyle {\sqrt {101}}-{\sqrt {100}}\approx 0.05}$ and ${\displaystyle {\sqrt {1000001}}-{\sqrt {1000000}}\approx 0.0005}$ (you may verify this yourself by taking the first-order Taylor approximation or a pocket calculator). We therefore suspect ${\displaystyle {\sqrt {k+1}}-{\sqrt {k}}}$ to be monotonously decreasing, which we will prove in the following:

${\displaystyle {\sqrt {k+1}}-{\sqrt {k}}={\frac {({\sqrt {k+1}}-{\sqrt {k}})({\sqrt {k+1}}+{\sqrt {k}})}{{\sqrt {k+1}}+{\sqrt {k}}}}={\frac {k+1-k}{{\sqrt {k+1}}+{\sqrt {k}}}}={\frac {1}{{\sqrt {k+1}}+{\sqrt {k}}}}}$

So

• ${\displaystyle a_{k}={\tfrac {1}{{\sqrt {k+1}}+{\sqrt {k}}}}}$ is indeed monotonously decreasing, since ${\displaystyle {\sqrt {k+2}}+{\sqrt {k+1}}\geq {\sqrt {k+1}}+{\sqrt {k}}\iff {\tfrac {1}{{\sqrt {k+2}}+{\sqrt {k+1}}}}\leq {\tfrac {1}{{\sqrt {k+1}}+{\sqrt {k}}}}\ \forall k\in \mathbb {N} }$
• ${\displaystyle (a_{k})}$ is a null sequence, since ${\displaystyle 0\leq a_{k}\leq {\tfrac {1}{2{\sqrt {k}}}}={\tfrac {1}{2}}{\tfrac {1}{\sqrt {k}}}\to 0}$.

So the series converges.

However, the series does not converge absolutely, since we can write ${\displaystyle \sum _{k}|a_{k}|}$ as a divergent telescoping series

${\displaystyle \sum _{k=1}^{\infty }({\sqrt {k+1}}-{\sqrt {k}})=\lim _{n\to \infty }\sum _{k=1}^{n}({\sqrt {k+1}}-{\sqrt {k}}){\underset {\text{summe}}{\overset {\text{Teleskop-}}{=}}}\lim _{n\to \infty }{\sqrt {n+1}}-1=\infty }$

7. Term test: For large ${\displaystyle k}$, the expression ${\displaystyle {\sqrt[{k}]{k}}}$ converges to 1. So we expect the absolute value of ${\displaystyle a_{k}={\frac {(-1)^{k}}{\sqrt[{k}]{k}}}}$ to converge to 1 and the series should diverge by means of the term test. Now, every second element ${\displaystyle a_{k}}$ is negative, so we have to restrict to a subsequence of positive elements:

${\displaystyle a_{k}={\frac {(-1)^{k}}{\sqrt[{k}]{k}}}\ \Rightarrow a_{2l}={\frac {1}{\sqrt[{2l}]{2l}}}\longrightarrow {\frac {1}{1}}=1{\text{ (da }}{\sqrt[{k}]{k}}\to 1{\text{)}}}$

Indeed, the subsequence ${\displaystyle (a_{2l})}$ of ${\displaystyle (a_{k})}$ doesn't converge to 0, so ${\displaystyle (a_{k})}$ cannot be a null sequence and by means of the term test, the series ${\displaystyle \sum _{k}a_{k}}$ is divergent.

Note: The ${\displaystyle (-1)^{k}}$ may tempt one to assume that the alternating series test is a good tool to use. However, the alternating series test can only prove convergence and as the series is divergent, any attempt using it bound to fail! Indeed, ${\displaystyle (b_{k})=({\tfrac {1}{\sqrt[{k}]{k}}})}$ is not a null sequence, which is why the alternating series test doesn't work.

8. alternating series test: The ${\displaystyle \left({\tfrac {k+1}{k}}\right)^{k}}$ converges to ${\displaystyle e}$, so the series elements behave like ${\displaystyle a_{k}\approx (-1)^{k}{\frac {e}{k}}}$ for large ${\displaystyle k}$. This is an alternating harmonic series, so we suspect ${\displaystyle \sum _{k}a_{k}}$ to converge, but not absolutely. We prove this using the alternating series test: For ${\displaystyle a_{k}={\frac {(1+{\tfrac {1}{k}})^{k}}{k}}=\left({\tfrac {k+1}{k}}\right)^{k}{\tfrac {1}{k}}}$ there is

• ${\displaystyle {\frac {a_{k+1}}{a_{k}}}={\frac {({\tfrac {k+2}{k+1}})^{k+1}{\tfrac {1}{k+1}}}{({\tfrac {k+1}{k}})^{k}{\tfrac {1}{k}}}}={\frac {({\tfrac {k+2}{k+1}})^{k+1}}{({\tfrac {k+1}{k}})^{k}{\tfrac {k+1}{k}}}}=\left({\frac {\tfrac {k+2}{k+1}}{\frac {k+1}{k}}}\right)^{k+1}=\left({\frac {(k+2)k}{(k+1)^{2}}}\right)^{k+1}=\left({\frac {k^{2}+2k}{k^{2}+2k+1}}\right)^{k+1}\leq 1}$ , so ${\displaystyle a_{k}}$ is monotonously decreasing.
• ${\displaystyle a_{k}={\frac {(1+{\tfrac {1}{k}})^{k}}{k}}\to 0}$, since ${\displaystyle (1+{\tfrac {1}{k}})^{k}\to e}$. So ${\displaystyle (a_{k})}$ is a null sequence.

Therefore, the series converges.

To show that it does not converge absolutely, we compare ${\displaystyle |a_{k}|}$ to a smaller but still divergent harmonic series (direct comparison test):

• ${\displaystyle |a_{k}|={\frac {(1+{\tfrac {1}{k}})^{k}}{k}}\geq {\frac {(1+{\tfrac {1}{1}})^{1}}{k}}={\frac {2}{k}}}$, since ${\displaystyle {\tilde {a}}_{k}=(1+{\tfrac {1}{k}})^{k}}$ is monotonously increasing.
• ${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {2}{k}}}$ still diverges. (harmonic series)

And hence, the series of absolute values ${\displaystyle \sum _{k=1}^{\infty }|a_{k}|=\sum _{k=1}^{\infty }{\frac {(1+{\tfrac {1}{k}})^{k}}{k}}}$ diverges.

Exercise (Convergence proof training 2)

Investigate, whether the following series are divergent, convergent or even absolutely convergent.

1. ${\displaystyle \sum _{k=2}^{\infty }{\frac {\ln k}{k^{3}}}}$
2. ${\displaystyle \sum _{k=2}^{\infty }{\frac {1}{\ln k}}}$
3. ${\displaystyle \sum _{k=1}^{\infty }k^{4}\exp(-k)}$
4. ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k+1}\cos({\tfrac {1}{k}})}$
5. ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k+1}\sin({\tfrac {1}{k}})}$
6. ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{\cosh k}}}$
7. ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k+1}{\frac {\sin k}{k^{2}}}}$

Solution (Convergence proof training 2)

1. Direct comparison test: It is useful to know that ${\displaystyle \ln k}$ eventually grows slower than any of the polynomials ${\displaystyle k,k^{0.5},k^{0.1},k^{0.01},...}$, i.e. any ${\displaystyle k^{\varepsilon }}$ with ${\displaystyle \varepsilon >0}$. We can generously take ${\displaystyle \epsilon =1}$ and estimate ${\displaystyle \ln k\leq k}$ :

• ${\displaystyle |a_{k}|={\frac {\ln k}{k^{3}}}\leq {\frac {k}{k^{3}}}={\frac {1}{k^{2}}}}$
• ${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {1}{k^{2}}}=\zeta (2)={\frac {\pi ^{2}}{6}}<\infty }$

So the series elements grow slower than ${\displaystyle {\frac {1}{k^{2}}}}$ and the series converges absolutely.

Note: For ${\displaystyle \varepsilon >0}$ , there is ${\displaystyle \ln k\leq c_{\varepsilon }k^{\varepsilon }}$ and we could even get ${\displaystyle |a_{k}|\leq {\frac {1}{k^{3-\varepsilon }}}}$.

2. Direct comparison test: Since ${\displaystyle \ln k\leq k}$, the series elements grow slower than

• ${\displaystyle a_{k}=|a_{k}|={\frac {1}{\ln k}}\geq {\frac {1}{k}}}$
• but ${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {1}{k}}}$ diverges (harmonic series)

So the series diverges.

3. Ratio test: We write ${\displaystyle a_{k}=k^{4}\exp(-k)={\tfrac {k^{4}}{e^{k}}}}$ . Both ${\displaystyle k^{4}}$ and ${\displaystyle e^{k}}$ grow fast at high ${\displaystyle k}$. But eventually, the exponential ${\displaystyle e^{k}}$ "wins" against all polynomial functions ,such that ${\displaystyle a_{k}\to 0}$ and even ${\displaystyle \sum _{k}a_{k}<\infty }$. Mathematically, this can be proven by considering the ratio between two subsequent elements:

${\displaystyle {\frac {|a_{k+1}|}{|a_{k}|}}={\frac {\tfrac {(k+1)^{4}}{e^{k+1}}}{\tfrac {k^{4}}{e^{k}}}}={\frac {(k+1)^{4}}{k^{4}}}{\frac {e^{k}}{e^{k+1}}}=\left(1+{\frac {1}{k}}\right)^{4}{\frac {1}{e}}\to {\frac {1}{e}}<1}$

So the series converges absolutely.

Note: We could have done the same proof with ${\displaystyle {\tilde {a}}_{k}={\tfrac {k^{n}}{e^{k}}}}$ for any ${\displaystyle n\geq 1}$ (not just ${\displaystyle n=4}$). The exponential always "wins" against the polynomial.

Alternative: Root est:

${\displaystyle {\sqrt[{k}]{|a_{k}|}}={\sqrt[{k}]{\tfrac {k^{4}}{e^{k}}}}={\frac {\sqrt[{k}]{k^{4}}}{\sqrt[{k}]{e^{k}}}}={\frac {{\sqrt[{k}]{k}}^{4}}{e}}\to {\frac {1}{e}}<1}$

So the series converges absolutely.

4. Term test: For ${\displaystyle a_{k}=(-1)^{k}\cos({\tfrac {1}{k}})}$ , the ${\displaystyle \cos({\tfrac {1}{k}})}$ approaches 1 as ${\displaystyle k\to \infty }$ . We select a subsequence consisting of only the positive entries:

${\displaystyle a_{2l}=\cos({\tfrac {1}{2l}})\to \cos(0)=1\neq 0}$

Therefore, ${\displaystyle (a_{k})}$ cannot be a null sequence and the corresponding series diverges.

Note: The ${\displaystyle (-1)^{k+1}}$ may trick you into taking the alternating series test for proving that the series converges. However, this doesn't work since ${\displaystyle b_{k}=\cos({\tfrac {1}{k}})}$ is not a null sequence!

5. Alternating series test: This time, ${\displaystyle \sin({\tfrac {1}{2l}})\to 0}$ , as ${\displaystyle k\to \infty }$ . So we have a ${\displaystyle (-1)^{k+1}}$ in front of a null sequence and may try the alternating series test:

• ${\displaystyle b_{k+1}=\sin({\tfrac {1}{k+1}})\leq \sin({\tfrac {1}{k}})=b_{k}}$, since ${\displaystyle \sin }$ is monotonously decreasing. Therefore, ${\displaystyle (b_{k})}$ is also monotonously decreasing.
• ${\displaystyle b_{k}=\sin({\tfrac {1}{k}})\to \sin(0)=0}$, since ${\displaystyle \sin }$ is continuous. Therefore, ${\displaystyle (b_{k})}$ is a null sequence.

By the alternating series test, the series converges.

Note: The series does not converge absolutely, since for large ${\displaystyle k}$, there is ${\displaystyle b_{k}=\sin({\tfrac {1}{k}})>{\tfrac {1}{2k}}}$. And ${\displaystyle {\tilde {b}}_{k}={\tfrac {1}{2k}}}$ is a divergent harmonic series.

6. Direct comparison test: The ${\displaystyle \cosh }$ can be written as ${\displaystyle a_{k}={\tfrac {1}{\cosh k}}={\tfrac {1}{\tfrac {e^{k}+e^{-k}}{2}}}={\tfrac {2}{e^{k}+e^{-k}}}}$. So it decays like ${\displaystyle {\frac {2}{e^{k}}}}$ (a geometric series). We compare ${\displaystyle a_{k}}$ to this geometric series

• ${\displaystyle |a_{k}|={\frac {2}{e^{k}+e^{-k}}}\leq {\tfrac {2}{e^{k}}}=2\cdot {\frac {1}{e^{k}}}}$, since ${\displaystyle e^{-k}>0}$.
• ${\displaystyle \sum _{k=1}^{\infty }2\cdot {\frac {1}{e^{k}}}=2\cdot \sum _{k=1}^{\infty }\left({\frac {1}{e}}\right)^{k}={\frac {2e}{1-{\tfrac {1}{e}}}}<\infty }$ (geometric series)

Therefore, the series converges absolutely.

7. Direct comparison test: The ${\displaystyle \sin }$ is oscillating with amplitude ${\displaystyle \leq 1}$ (therefore staying bounded) and the ${\displaystyle {\frac {1}{k^{2}}}}$ decays fast enough to get convergence. We can hence compare:

• ${\displaystyle |a_{k}|={\frac {|\sin k|}{k^{2}}}\leq {\tfrac {1}{k^{2}}}}$
• ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k^{2}}}=\zeta (2)={\frac {\pi ^{2}}{6}}<\infty }$

So the series converges absolutely.

Note: Even though the series is oscillating, the alternating series test doesn't work here, since ${\displaystyle b_{k}={\tfrac {\sin k}{k^{2}}}}$ is not monotonously decreasing!

Exercise (Series depending on parameters)

The following series depend on a parameter ${\displaystyle x\in R}$ . For some ${\displaystyle x}$ , they will converge, for others not. Your job is to investigate for which ${\displaystyle x\in \mathbb {R} }$ , the series converges (absolutely) or diverges:

1. ${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {x^{k}}{(2k)!}}}$
2. ${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {x^{k}}{k^{2}}}}$
3. ${\displaystyle \sum \limits _{k=1}^{\infty }{\sqrt {k}}x^{k}}$
4. ${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {x^{k}}{k}}}$

Solution (Series depending on parameters)

1. Ratio test: Intuitively, a factorial ${\displaystyle k!}$ grows faster than any exponential ${\displaystyle x^{k}}$ in a way that not just ${\displaystyle {\frac {x^{k}}{k!}}\to 0}$ , but even the series ${\displaystyle \sum _{k}{\frac {x^{k}}{k!}}}$ converges. This should also hold true if we replace ${\displaystyle k!}$ by the larger expression ${\displaystyle (2k)!}$ . We can verify this mathematically by the ratio test:

${\displaystyle {\frac {|a_{k+1}|}{|a_{k}|}}={\frac {\frac {|x|^{k+1}}{(2k+2)!}}{\frac {|x|^{k}}{(2k)!}}}={\frac {|x|^{k+1}}{|x|^{k}}}\cdot {\frac {(2k)!}{(2k+2)!}}={\frac {|x|}{(2k+1)(2k+2)}}={\frac {|x|}{4k^{2}+6k+2}}\to 0<1}$

So series is absolutely convergent for all ${\displaystyle x\in \mathbb {R} }$.

2. Case distinction:

Fall 1: ${\displaystyle |x|\leq 1}$

The series elements are smaller than ${\displaystyle |a_{k}|={\frac {|x|^{k}}{k^{2}}}\leq {\frac {1}{k^{2}}}}$ , which is known to be a convergent series: ${\displaystyle \sum \limits _{n=1}^{\infty }{\frac {1}{k^{2}}}=\zeta (2)={\tfrac {\pi ^{2}}{6}}<\infty }$

So by the direct comparison test, the series converges absolutely.

Fall 2: ${\displaystyle |x|>1}$

In this case, we expect the ${\displaystyle x^{k}}$ to "explode exponentially" in ${\displaystyle x}$, while ${\displaystyle k^{2}}$ is growing only polynomially (and hence slower). So the series ${\displaystyle {\frac {x^{k}}{k^{2}}}}$ should converge. Mathematically, this can be verified by the root test: ${\displaystyle {\sqrt[{k}]{|a_{k}|}}={\sqrt[{k}]{\frac {|x|^{k}}{k^{2}}}}={\frac {|x|}{\sqrt[{k}]{k^{2}}}}={\frac {|x|}{{\sqrt[{k}]{k}}^{2}}}\to |x|>1}$, da ${\displaystyle {\sqrt[{k}]{k}}\to 1}$

So the series diverges.

3. Case distinction:

Wir unterscheiden zwei Fälle:

Fall 1: ${\displaystyle |x|<1}$

In that case, the ${\displaystyle x^{k}}$ decays exponentially, and therefore "wins" against the polynomially increasing ${\displaystyle {\sqrt {k}}=k^{1/2}}$. We use the ratio test to verify this:

${\displaystyle {\frac {|a_{k+1}|}{|a_{k}|}}={\frac {{\sqrt {k+1}}|x|^{k+1}}{{\sqrt {k}}|x|^{k}}}={\sqrt {1+{\frac {1}{k}}}}|x|\to |x|<1}$

So indeed, the series diverges absolutely.

Fall 2: ${\displaystyle |x|\geq 1}$

Here, the ${\displaystyle x^{k}}$ is constant (${\displaystyle |x|=1}$) or exponentially increasing. Multiplying it by ${\displaystyle {\sqrt {k}}}$ we get sequence elements which are increasing and hence bigger than a constant. This suggests using the term test: ${\displaystyle |a_{k}|={\sqrt {k}}|x|^{k}\geq {\sqrt {k}}\to \infty }$. So ${\displaystyle (a_{k})}$ is not a null sequence.

Therefore, the series diverges.

4. Case distinction:

Fall 1: ${\displaystyle |x|<1}$

Here, ${\displaystyle |a_{k}|={\frac {|x|^{k}}{k}}\leq |x|^{k}}$ and ${\displaystyle \sum \limits _{n=1}^{\infty }|x|^{k}<\infty }$ (geometric series)

So by the direct comparison test, the series converges absolutely.

Fall 2: ${\displaystyle x=1}$

${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {1^{k}}{k}}=\sum \limits _{k=1}^{\infty }{\frac {1}{k}}}$ , which is a divergent harmonic series.

Fall 3: ${\displaystyle x=-1}$

${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {(-1)^{k}}{k}}}$ is an alternating harmonic series, which converges but not absolutely, by the alternating series test.

Fall 4: ${\displaystyle |x|>1}$

We have ${\displaystyle |a_{k}|={\frac {|x|^{k}}{k}}\geq {\frac {1}{k}}}$, where the geometric series ${\displaystyle \sum \limits _{k=1}^{\infty }{\frac {1}{k}}}$ diverges (direct comparison test).

Note: The cases ${\displaystyle |x|<1}$ and ${\displaystyle |x|>1}$ can also be treated by the root test or the ratio test.

Exercise (Direct comparison in asymptotic behaviour)

Investigate, whether the series

${\displaystyle \sum _{k=1}^{\infty }{\frac {7k^{7}(1+{\frac {1}{k}})(k^{3}-1)}{(k^{3}+2)(2k^{5}+1)(k^{2}+8)^{2}}}}$

converges or diverges.

Solution (Direct comparison in asymptotic behaviour)

We take a look at the asymptotic behaviour for ${\displaystyle k\to \infty }$:

${\displaystyle a_{k}={\frac {7k^{10}(1+{\frac {1}{k}})(1-{\frac {1}{k^{3}}})}{2k^{12}(1+{\frac {2}{k^{3}}})(1+{\frac {1}{2k^{5}}})(1+{\frac {8}{k^{2}}})^{2}}}=\underbrace {\frac {7k^{10}}{2k^{12}}} _{={\frac {7}{2}}{\frac {1}{k^{2}}}}\underbrace {\frac {(1+{\frac {1}{k}})(1-{\frac {1}{k^{3}}})}{(1+{\frac {2}{k^{3}}})(1+{\frac {1}{2k^{5}}})(1+{\frac {8}{k^{2}}})^{2}}} _{\rightarrow 1{\text{ for }}k\to \infty }}$

So it might be useful to compare ${\displaystyle a_{k}}$ with a series scaling like ${\displaystyle b_{k}={\tfrac {1}{k^{2}}}}$. In fact,

${\displaystyle \lim _{k\to \infty }{\frac {a_{k}}{b_{k}}}=\lim _{k\to \infty }{\frac {7}{2}}{\frac {(1+{\frac {1}{k}})(1-{\frac {1}{k^{3}}})}{(1+{\frac {2}{k^{3}}})(1+{\frac {1}{2k^{5}}})(1+{\frac {8}{k^{2}}})^{2}}}={\frac {7}{2}}{\frac {(1+0)(1-0)}{(1+0)(1+0)(1+0)^{2}}}={\frac {7}{2}}}$

This convergence tells us that for high ${\displaystyle k}$ , there must be a ${\displaystyle M\in \mathbb {N} }$ with

${\displaystyle |a_{k}|\leq 2\cdot {\frac {7}{2}}|b_{k}|=7b_{k}}$ for all ${\displaystyle k\geq M}$

So we can compare ${\displaystyle |a_{k}|}$ to ${\displaystyle 7b_{k}={\tfrac {7}{k^{2}}}}$ . Since the series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {7}{k^{2}}}=7{\tfrac {\pi ^{2}}{6}}}$converges, we know by the direct comparison test that also ${\displaystyle \sum _{k=1}^{\infty }}$ converges absolutely.

## Term test - strengthened version

Exercise (Term test - strengthened version)

The term test tells us that if ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is monotonously decreasing and ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ converges, then ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is a null sequence. Prove that in this case, even ${\displaystyle (na_{n})_{n\in \mathbb {N} }}$ must be a null sequence. (This means that it suffices to show that ${\displaystyle (na_{n})_{n\in \mathbb {N} }}$ is not a null sequence in order to conclude that ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ diverges.)

Solution (Term test - strengthened version)

The intuition behind this test is that ${\displaystyle (na_{n})_{n\in \mathbb {N} }}$ being not a null sequence means that ${\displaystyle (a_{n})}$ decays slower than or as fast as a harmonic sequence ${\displaystyle b_{n}={\tfrac {1}{n}}}$, and ${\displaystyle \sum _{n=1}^{\infty }b_{n}}$ diverges. A formal proof can be done by using the Cauchy criterion: Assume that ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ converges. then the sequence of partial sums ${\displaystyle (s_{k})_{k\in \mathbb {N} }=\sum _{n=1}^{k}a_{n}}$ is a Cauchy sequence. We will first use this fact to bound all even elements of ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$:

Proof step: ${\displaystyle (2ka_{2k})_{k\in \mathbb {N} }}$ is a null sequence

Since ${\displaystyle (s_{k})_{k\in \mathbb {N} }=\sum _{n=1}^{k}a_{n}}$ is a Cauchy sequence, there must be an ${\displaystyle \epsilon >0}$ and an ${\displaystyle N\in \mathbb {N} }$, such that for all ${\displaystyle k+1\geq N}$ there is

${\displaystyle |s_{n}-s_{k}|=\sum _{n=k+1}^{2k}a_{n}=a_{k+1}+a_{k+2}+\ldots +a_{2k-1}+a_{2k}<\epsilon }$

Hence, for all ${\displaystyle k+1\geq N}$:

{\displaystyle {\begin{aligned}{\frac {1}{2}}\cdot 2ka_{2k}&=ka_{2k}\\[0.3em]&=a_{2k}+a_{2k}+\ldots +a_{2k}+a_{2k}\\[0.3em]&\left\downarrow \ {\color {grey}(a_{n}){\text{ decreases monotonously}}}\right.\\[0.3em]&\leq a_{k+1}+a_{k+2}+\ldots +a_{2k-1}+a_{2k}<\epsilon \\[0.3em]\end{aligned}}}

So we know that ${\displaystyle (ka_{2k})}$ and hence also ${\displaystyle (2ka_{2k})}$ eventually falls below each ${\displaystyle \epsilon >0}$ , so they are null sequences.

Now we are still missing the odd numbers:

Proof step: ${\displaystyle ((2k-1)a_{2k-1})_{k\in \mathbb {N} }}$ is a null sequence

The argumentation is the same as above: ${\displaystyle (s_{k})_{k\in \mathbb {N} }=\sum _{n=1}^{k}a_{n}}$ is a Cauchy sequence, so for each ${\displaystyle \epsilon >0}$ there is an ${\displaystyle N\in \mathbb {N} }$, such that for all ${\displaystyle k\geq N}$ there is

${\displaystyle |s_{2k-1}-s_{k-1}|\sum _{n=k}^{2k-1}a_{n}=a_{k}+a_{k+1}+\ldots +a_{2k-2}+a_{2k-1}<\epsilon }$

Hence, for all ${\displaystyle k\geq N}$:

{\displaystyle {\begin{aligned}{\frac {1}{2}}\cdot (2k-1)a_{2k-1}&\leq {\frac {1}{2}}\cdot 2ka_{2k-1}\\[0.3em]&=ka_{2k-1}\\[0.3em]&=a_{2k-1}+a_{2k-1}+\ldots +a_{2k-1}+a_{2k-1}\\[0.3em]&\left\downarrow \ {\color {grey}(a_{n}){\text{ decreases monotonously}}}\right.\\[0.3em]&\leq a_{k}+a_{k+1}+\ldots +a_{2k-2}+a_{2k-1}<\epsilon \\[0.3em]\end{aligned}}}

So we know that ${\displaystyle ({\tfrac {1}{2}}\cdot (2k-1)a_{2k-1})}$ and hence also ${\displaystyle ((2k-1)a_{2k-1})}$ are null sequences.

The case distinction was actually only necessary since for even numbers ${\displaystyle K=2k}$, we had to sum over ${\displaystyle k={\tfrac {K}{2}}}$ and for odd numbers ${\displaystyle K=2k-1}$ over ${\displaystyle k={\tfrac {K+1}{2}}}$ elements. Since both for even ${\displaystyle (2ka_{2k})}$ and odd indices ${\displaystyle ((2k-1)a_{2k-1})}$ , the elements tend to zero as ${\displaystyle k\to \infty }$, the sequence ${\displaystyle (na_{n})}$ must also tend to zero, i.e. it is a null sequence.

## Cauchy criterion: an application

Exercise (Alternating harmonic series)

Using the Cauchy criterion, prove that the alternating harmonic series ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}}$ converges. (note: the harmonic series ${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k}}}$ does not converge!)

Solution (Alternating harmonic series)

Our aim is to show that the sequence of partial sums ${\displaystyle s_{n}=\sum _{k=1}^{n}(-1)^{k+1}{\frac {1}{k}}}$ is a Cauchy sequence. That means, we have to put a bound on

{\displaystyle {\begin{aligned}|s_{n}-s_{m-1}|=\left|\sum _{k=m}^{n}(-1)^{k+1}{\frac {1}{k}}\right|&=\left|(-1)^{m+1}{\frac {1}{m}}+(-1)^{m+2}{\frac {1}{m+1}}+(-1)^{m+3}{\frac {1}{m+2}}+(-1)^{m+4}{\frac {1}{m+3}}+(-1)^{m+5}{\frac {1}{m+4}}+\ldots \right|\\[0.5em]&=\underbrace {\left|(-1)^{m+1}\right|} _{=1}|\overbrace {{\frac {1}{m}}\underbrace {-({\frac {1}{m+1}}-{\frac {1}{m+2}})} _{\leq 0}\underbrace {-({\frac {1}{m+3}}-{\frac {1}{m+4}})} _{\leq 0}\underbrace {-\ldots } _{\leq 0}} ^{\geq 0}|\\[0.5em]&\leq \left|{\frac {1}{m}}\right|\\[0.5em]&={\frac {1}{m}}\end{aligned}}}

The sequence ${\displaystyle ({\tfrac {1}{m}})_{m\in \mathbb {N} }}$ is definitely null. So for each ${\displaystyle \epsilon >0}$ we can choose an ${\displaystyle N\in \mathbb {N} }$, such that ${\displaystyle |s_{n}-s_{m-1}|=\left|\sum _{k=m}^{n}(-1)^{k+1}{\tfrac {1}{k}}\right|\leq |{\tfrac {1}{m}}|<\epsilon }$ for all ${\displaystyle n\geq m\geq N}$. So ${\displaystyle (s_{n})_{n\in \mathbb {N} }}$ is a Cauchy sequence implying that ${\displaystyle \lim _{n\to \infty }s_{n}=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}}$ converges, which was to be shown.

## Root and ratio test: estimate of errors

In some cases, the infinite sum ${\displaystyle s=\sum _{k=1}^{\infty }a_{k}}$ cannot be computed explicitly. We can try to approximate it by ${\displaystyle s_{n}=\sum _{k=1}^{n}a_{k}}$. The following exercises are devoted to giving upper bounds on the approximation error ${\displaystyle |s-s_{n}|}$:

Exercise (Error estimate for the root test)

Let ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ be a sequence and ${\displaystyle q<1}$. Further, ${\displaystyle {\sqrt[{n}]{|a_{n}|}}\leq q}$ for all ${\displaystyle n\geq N}$. Then, the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}=s}$ converges absolutely by the root test. Show that it can be approximated by ${\displaystyle s_{n}=\sum _{k=1}^{n}a_{k}}$ with ${\displaystyle n\geq N-1}$ up to a maximal error of

${\displaystyle \left|s-s_{n}\right|\leq {\frac {q^{n+1}}{1-q}}}$

Solution (Error estimate for the root test)

The idea is that the root test bounds ${\displaystyle \sum _{k}|a_{k}|}$ by a geometric series ${\displaystyle \sum _{k}q^{k}}$ and ${\displaystyle \sum _{k=n+1}^{\infty }q^{k}={\frac {q^{n+1}}{1-q}}}$ This bounding can be explicitly shown by using the root test criterion:

${\displaystyle {\sqrt[{k}]{|a_{k}|}}\leq q\iff |a_{k}|\leq q^{k}}$

So for ${\displaystyle n\geq N-1\iff n+1\geq N}$, we can sum up:

{\displaystyle {\begin{aligned}\left|s-s_{n}\right|&=\left|\sum _{n+1}^{\infty }a_{k}\right|\\[0.3em]&{\color {grey}\left\downarrow \ {\text{generalized triangle inequality}}\right.}\\[0.3em]&\leq \sum _{k=n+1}^{\infty }|a_{k}|\\[0.3em]&\leq \sum _{k=n+1}^{\infty }q^{k}\\[0.3em]&{\color {grey}\left\downarrow \ {\text{index shift}}\right.}\\[0.3em]&=\sum _{k=0}^{\infty }q^{n+1+k}\\[0.3em]&=\sum _{k=0}^{\infty }q^{n+1}\cdot q^{k}\\[0.3em]&{\color {grey}\left\downarrow \ {\text{factorization rule}}\right.}\\[0.3em]&=q^{n+1}\cdot \sum _{k=0}^{\infty }q^{k}\\[0.3em]&{\color {grey}\left\downarrow \ {\text{geometric series}}\right.}\\[0.3em]&=q^{n+1}\cdot {\frac {1}{1-q}}={\frac {q^{n+1}}{1-q}}\end{aligned}}}

Exercise (Error estimate for the ratio test)

Let ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ be a sequence and ${\displaystyle q<1}$. Further, let ${\displaystyle a_{n}\neq 0}$ and ${\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|\leq q}$ for all ${\displaystyle n\geq N}$. By the ratio test, the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}=s}$ converges absolutely. Show that it can be approximated by ${\displaystyle s_{n}=\sum _{k=1}^{n}a_{k}}$ with ${\displaystyle n\geq N-1}$ up to a maximal error of

${\displaystyle \left|s-s_{n}\right|\leq {\frac {|a_{n+1}|}{1-q}}}$

Solution (Error estimate for the ratio test)

Again, we bound the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ by a geometric series. But this time, the factor of ${\displaystyle q^{n+1}}$ is replaced by ${\displaystyle |a_{n+1}|}$. By assumption:

${\displaystyle \left|{\frac {a_{k+1}}{a_{k}}}\right|\leq q\iff |a_{k+1}|\leq q\cdot |a_{k}|}$

For any ${\displaystyle n\geq N-1\iff n+1\geq N}$, we can conclude by iteration:

{\displaystyle {\begin{aligned}&|a_{n+2}|\leq q\cdot |a_{n+1}|\\[0.5em]\Longrightarrow &|a_{n+3}|\leq q\cdot |a_{n+2}|\leq q^{2}\cdot |a_{n+1}|\\[0.5em]\Longrightarrow &|a_{n+4}|\leq q^{3}\cdot |a_{n+1}|\\[0.5em]&{\color {grey}\left\downarrow \ (k-3){\text{-fold iteration of the inequality}}\right.}\\[0.5em]\Longrightarrow &|a_{n+1+k}|\leq q^{k}\cdot |a_{n+1}|\end{aligned}}}

Again, we sum up the geometric series bounds:

{\displaystyle {\begin{aligned}\left|s-s_{n}\right|&=\left|\sum _{n+1}^{\infty }a_{k}\right|\\[0.3em]&{\color {grey}\left\downarrow \ {\text{generalized triangle inequality}}\right.}\\[0.3em]&\leq \sum _{k=n+1}^{\infty }|a_{k}|\\[0.3em]&{\color {grey}\left\downarrow \ {\text{index shift}}\right.}\\[0.3em]&=\sum _{k=0}^{\infty }|a_{n+1+k}|\\[0.3em]&\leq \sum _{k=0}^{\infty }q^{k}\cdot |a_{n+1}|\\[0.3em]&{\color {grey}\left\downarrow \ {\text{factorization rule}}\right.}\\[0.3em]&=|a_{n+1}|\cdot \sum _{k=0}^{\infty }q^{k}\\[0.3em]&{\color {grey}\left\downarrow \ {\text{geometric series}}\right.}\\[0.3em]&=|a_{n+1}|\cdot {\frac {1}{1-q}}={\frac {|a_{n+1}|}{1-q}}\end{aligned}}}

Exercise (showing that a sequence is null)

1. Let ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ be a sequence and ${\displaystyle 0. Further, let ${\displaystyle a_{n}\neq 0}$ and ${\displaystyle \lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=q}$ or ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{|a_{n}|}}=q}$. Show that we then have a null sequence ${\displaystyle \lim _{n\to \infty }a_{n}=0}$.
2. Conclude that ${\displaystyle \lim _{n\to \infty }{\binom {n}{k}}q^{n}=0}$ for ${\displaystyle 0 and ${\displaystyle k\in \mathbb {N} _{0}}$.

Solution (showing that a sequence is null)

1. Both ${\displaystyle \lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=q<1}$ (ratio test) and ${\displaystyle \lim _{n\to \infty }{\sqrt[{n}]{|a_{n}|}}=q<1}$ (root test) imply that the series ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ converges. But by means of the term test , the sequence ${\displaystyle \lim _{n\to \infty }a_{n}=0}$ has to be a null sequence.
2. We specifically consider the sequence ${\displaystyle a_{n}={\binom {n}{k}}q^{n}}$ . It satisfies the ratio test criterion:
{\displaystyle {\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }{\frac {{\binom {n+1}{k}}q^{n+1}}{{\binom {n}{k}}q^{n}}}\\[0.3em]&=\lim _{n\to \infty }{\frac {{\frac {(n+1)n(n-1)\cdot \ldots \cdot ((n+1)-k+1)}{k!}}\cdot q\cdot q^{n}}{{\frac {n(n-1)\cdot \ldots \cdot (n-k+1)}{k!}}\cdot q^{n}}}\\[0.3em]&=\lim _{n\to \infty }{\frac {(n+1)n(n-1)\cdot \ldots \cdot (n-k+2)}{n(n-1)\cdot \ldots \cdot (n-k+2)(n-k+1)}}\cdot q\\[0.3em]&=\lim _{n\to \infty }{\frac {n+1}{n-k+1}}\cdot q\\[0.3em]&=\lim _{n\to \infty }{\frac {1+\overbrace {\frac {1}{n}} ^{\to 0}}{1-\underbrace {\frac {k}{n}} _{\to 0}+\underbrace {\frac {1}{n}} _{\to 0}}}\cdot q\\[0.3em]&=q<1\end{aligned}}}

So the series ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ converges and by part 1 of the exercise, ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ is a null sequence, i.e. ${\displaystyle \lim _{n\to \infty }{\binom {n}{k}}q^{n}=0}$.

## Alternating series test with error bounds

Exercise (Alternating series test with error bounds)

First, show that the series

${\displaystyle \sum _{k=1}^{\infty }{\frac {(-1)^{k+1}}{2k+1}}{\frac {k}{k+2}}=:\sum _{k=1}^{\infty }a_{k}=:S}$

converges. Then, determine an index ${\displaystyle n_{0}\in \mathbb {N} }$, such that the partial sum ${\displaystyle S_{n}=\sum _{k=1}^{n}{\frac {(-1)^{k+1}}{2k+1}}{\frac {k}{k+2}}}$ for ${\displaystyle n\geq n_{0}}$ approximates ${\displaystyle S}$ up to a precision of ${\displaystyle |S-S_{N}|}$\tfrac{1}{100}[/itex] .

Solution (Alternating series test with error bounds)

Proof step: The series converges

For ${\displaystyle b_{k}=|a_{k}|={\tfrac {k}{(2k+1)(k+2)}}}$ we can estimate

{\displaystyle {\begin{aligned}{\frac {b_{k+1}}{b_{k}}}&={\frac {\frac {k+1}{(2k+3)(k+3)}}{\frac {k}{(2k+1)(k+2)}}}\\[0.5em]&={\frac {(k+1)(2k+1)(k+2)}{k(2k+3)(k+3)}}\\[0.5em]&={\frac {2k^{3}+7k^{2}+7k+2}{2k^{3}+9k^{2}+9k}}\\[0.5em]&={\frac {2k^{3}+7k^{2}+7k+2}{2k^{3}+7k^{2}+7k+2k^{2}+2k}}\\[0.5em]&={\frac {2k^{3}+7k^{2}+7k+2}{2k^{3}+7k^{2}+7k+2(k^{2}+k)}}\\[0.5em]&\left\downarrow \ k^{2}+k\geq 1\right.\\[0.5em]&\leq {\frac {2k^{3}+7k^{2}+7k+2}{2k^{3}+7k^{2}+7k+2}}\\[0.5em]&=1\end{aligned}}}

So ${\displaystyle (b_{k})_{k\in \mathbb {N} }}$ is monotonously decreasing. Further

{\displaystyle {\begin{aligned}\lim _{k\to \infty }b_{k}&=\lim _{k\to \infty }{\frac {k}{(2k+1)(k+2)}}\\[0.5em]&=\lim _{k\to \infty }{\frac {k}{2k^{2}+5k+2}}\\[0.5em]&=\lim _{k\to \infty }{\frac {\frac {1}{k}}{2+{\frac {5}{k}}+{\frac {2}{k^{2}}}}}\\[0.5em]&\left\downarrow \ {\text{calculus for sequences}}\right.\\[0.5em]&\leq {\frac {0}{2+0+0}}\\[0.5em]&=0\end{aligned}}}

So ${\displaystyle (b_{k})}$ is even a null sequence. By means of the alternating series test, the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ hence converges.

Proof step: Finding ${\displaystyle n_{0}}$

For the alternating series test, there is an error bound:

${\displaystyle \left|\sum _{k=1}^{\infty }b_{k}-\sum _{k=1}^{n}b_{k}\right|

With ${\displaystyle b_{n+1}={\frac {n+1}{(2n+3)(n+3)}}}$. It saves a ot of calculation work to just give a coarse upper bound for this expression:

${\displaystyle b_{n+1}={\frac {n+1}{(2n+3)(n+3)}}\leq {\frac {n+3}{(2n+3)(n+3)}}={\frac {1}{2n+3}}}$

Now, if ${\displaystyle {\frac {1}{2n+3}}<{\tfrac {1}{100}}}$, then there is also ${\displaystyle b_{n+1}<{\tfrac {1}{100}}}$. Hence,

{\displaystyle {\begin{aligned}{\frac {1}{2n+3}}<{\frac {1}{100}}&\iff 2n+3>100\\[0.5em]&\iff 2n>97\\[0.5em]&\iff n>49\in \mathbb {N} \end{aligned}}}

So ${\displaystyle n+1=49}$ (and any bigger ${\displaystyle n}$) leads to a precise enough approximation. For ${\displaystyle n_{0}=n=48}$ the approximation precision is therefore better than ${\displaystyle |S-S_{N}|\leq {\tfrac {1}{100}}}$ .

## Cauchy condensation test

Exercise (A double logarithm)

Determine, for which ${\displaystyle \alpha >0}$ the following series converges:

${\displaystyle \sum _{k=1}^{\infty }{\frac {1}{k\ln(k)\ln(\ln(k))^{\alpha }}}}$

Solution (A double logarithm)

The Cauchy condensation test is useful to resolve the double logarithm: if in ${\displaystyle {\tfrac {1}{\ln(\ln(n))}}}$ we only consider the elements ${\displaystyle n=2^{k}}$ (i.e. 2, 4, 8, 16, ...), we get obtain a single logarithm in ${\displaystyle k}$ instead of a double logarithm: ${\displaystyle {\tfrac {1}{\ln(\ln(n))}}={\tfrac {1}{\ln(\ln(2^{k}))}}={\tfrac {1}{\ln(k\ln 2)}}}$. As in our sequence, ${\displaystyle a_{k}={\tfrac {1}{k\ln(k)\ln(\ln(k))^{\alpha }}}}$ is monotonously decreasing, the Cauchy condensation test yields that it converges if and only if the following series converges:

{\displaystyle {\begin{aligned}\sum \limits _{k=1}^{\infty }2^{k}{\frac {1}{2^{k}\ln(2^{k})\ln(\ln(2^{k}))^{\alpha }}}&=\sum \limits _{k=1}^{\infty }{\frac {1}{\ln(2^{k})\ln(\ln(2^{k}))^{\alpha }}}\\[0.5em]&\ {\color {Gray}\left\downarrow \ \ln(x^{y})=y\ln(x)\right.}\\[0.5em]&=\sum \limits _{k=1}^{\infty }{\frac {1}{k\ln(2)\ln(k\ln(2))^{\alpha }}}\\[0.5em]&\ {\color {Gray}\left\downarrow \ \ln(xy)=\ln(x)+\ln(y)\right.}\\[0.5em]&=\sum \limits _{k=1}^{\infty }{\frac {1}{k\ln(2)(\ln(k)+\ln(\ln(2)))^{\alpha }}}\end{aligned}}}

In the article "Cauchy condensation test", there is an exercise proving that ${\displaystyle \sum \limits _{n=2}^{\infty }{\frac {1}{n(\ln(n))^{\alpha }}}}$ converges for ${\displaystyle \alpha >1}$ and diverges for ${\displaystyle 0<\alpha \leq 1}$. This can be used for a direct comparison:

Fall 1: ${\displaystyle \alpha >1}$

Here,

${\displaystyle {\frac {1}{k\ln(2)(\ln(k)+\ln(\ln(2)))^{\alpha }}}\leq {\frac {1}{k\ln(2)\ln(k)^{\alpha }}}={\frac {1}{\ln(2)}}{\frac {1}{k\ln(k)^{\alpha }}}}$

and ${\displaystyle {\frac {1}{\ln(2)}}\sum \limits _{k=1}^{\infty }{\frac {1}{k\ln(k)^{\alpha }}}<\infty }$. So by direct comparison, our series converges for all ${\displaystyle \alpha >1}$.

Fall 2: ${\displaystyle 0<\alpha \leq 1}$

Here,

${\displaystyle {\frac {1}{k\ln(2)(\ln(k)+\ln(\ln(2)))^{\alpha }}}\geq {\frac {1}{k\ln(2)(\ln(k)+\ln(k))^{\alpha }}}={\frac {1}{k\ln(2)(2\ln(k))^{\alpha }}}={\frac {1}{k\ln(2)2^{\alpha }\ln(k)^{\alpha }}}={\frac {1}{2^{\alpha }\ln(2)}}{\frac {1}{k\ln(k)^{\alpha }}}}$

and ${\displaystyle {\frac {1}{2^{\alpha }\ln(2)}}\sum \limits _{k=1}^{\infty }{\frac {1}{k\ln(k)^{\alpha }}}}$ diverges. So by direct comparison, the series diverges for all ${\displaystyle 0<\alpha \leq 1}$.

## Further convergence criteria

Exercise (Series with products)

Let ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ and ${\displaystyle (b_{k})_{k\in \mathbb {N} }}$ be two real sequences. Prove:

1. Whenever the series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ converges absolutely and ${\displaystyle (b_{k})_{k\in \mathbb {N} }}$ is bounded, then also ${\displaystyle \sum _{k=1}^{\infty }a_{k}b_{k}}$ converges absolutely.
2. There are pairs of a series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ , which converges and a sequence ${\displaystyle (b_{k})_{k\in \mathbb {N} }}$ which is bounded, such that ${\displaystyle \sum _{k=1}^{\infty }a_{k}b_{k}}$ does not converge. So the above statement does only hold true for absolute convergence, but not for convergence in general.

Solution (Series with products)

Part 1:

Way 1: Boundedness of partial sums

Since ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ converges absolutely, the sequence of partial sums ${\displaystyle \left(\sum _{k=1}^{n}|a_{k}|\right)_{n\in \mathbb {N} }}$ must be bounded. Further, ${\displaystyle (b_{k})_{k\in \mathbb {N} }}$ is bounded. That means, there is an upper bound ${\displaystyle S>0}$ with ${\displaystyle |b_{k}|\leq S}$ for all ${\displaystyle k\in \mathbb {N} }$ and therefore

${\displaystyle \sum _{k=1}^{n}|a_{k}b_{k}|\leq \sum _{k=1}^{n}|a_{k}|\cdot S=S\cdot \sum _{k=1}^{n}|a_{k}|}$

But now, ${\displaystyle \left(\sum _{k=1}^{n}|a_{k}|\right)_{n\in \mathbb {N} }}$ is bounded, so ${\displaystyle \left(S\cdot \sum _{k=1}^{n}|a_{k}|\right)_{n\in \mathbb {N} }}$ is bounded as well. The sequence ${\displaystyle \left(\sum _{k=1}^{n}|a_{k}b_{k}|\right)_{n\in \mathbb {N} }}$ is smaller than ${\displaystyle \left(S\cdot \sum _{k=1}^{n}|a_{k}|\right)_{n\in \mathbb {N} }}$ and therefore also bounded, i.e. ${\displaystyle \sum _{k=1}^{\infty }a_{k}b_{k}}$ converges absolutely.

Way 2: Direct comparison test

As above, ${\displaystyle (b_{k})_{k\in \mathbb {N} }}$ has an upper bound ${\displaystyle S>0}$ , meaning that ${\displaystyle |b_{k}|\leq S}$ for all ${\displaystyle k\in \mathbb {N} }$. Hence,

${\displaystyle |a_{k}b_{k}|=|a_{k}|\cdot |b_{k}|\leq S\cdot |a_{k}|}$

Now, the series ${\displaystyle \sum _{k=1}^{\infty }|a_{k}|}$ converges, as well as ${\displaystyle S\cdot \sum _{k=1}^{\infty }|a_{k}|}$ . We use the direct comparison test: ${\displaystyle \sum _{k=1}^{\infty }|a_{k}b_{k}|}$ is smaller than ${\displaystyle S\cdot \sum _{k=1}^{\infty }|a_{k}|}$ and hence converges.

Part 2: We know that the harmonic series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{k}}}$ diverges, but can be made convergent by putting a minus sign in front of every second element ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {(-1)^{k}}{k}}<\infty }$ . The idea is now to "reverse" this process: we take the convergent alternating series ${\displaystyle a_{k}={\tfrac {(-1)^{k}}{k}}}$ and make it again divergent turning every second element positive:

${\displaystyle \underbrace {\frac {(-1)^{k}}{k}} _{=a_{k}}\underbrace {(-1)^{k}} _{=b_{k}}=(-1)^{2k}{\frac {1}{k}}={\frac {1}{k}}}$

The "reversing sequence" ${\displaystyle (b_{k})_{k\in \mathbb {N} }}$ is bounded, as ${\displaystyle |b_{k}|=|(-1)^{k}|=1\leq 1=S}$. So ${\displaystyle \sum _{k=1}^{\infty }a_{k}=\sum _{k=1}^{\infty }{\tfrac {(-1)^{k}}{k}}}$ converges and ${\displaystyle (b_{k})_{k\in \mathbb {N} }=((-1)^{k})_{k\in \mathbb {N} }}$ is bounded, but ${\displaystyle \sum _{k=1}^{\infty }a_{k}b_{k}=\sum _{k=1}^{\infty }{\tfrac {1}{k}}}$ diverges as a harmonic series.

Note: other counter-examples are also possible.

Exercise (Raabe criterion)

1. Let ${\displaystyle (a_{n})_{n\in \mathbb {N} }}$ and ${\displaystyle (b_{n})_{n\in \mathbb {N} }}$ be two real sequences. Prove: If for almost all ${\displaystyle n\in \mathbb {N} }$ , there is ${\displaystyle a_{n}\neq 0}$ (all except for a finite number of elements). Then, if
• ${\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|\leq 1-{\frac {c}{n+1}}}$ for some ${\displaystyle c>1}$, then ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ converges absolutely.
• ${\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|\geq 1-{\frac {1}{n+1}}}$ then ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ diverges.
2. Use this so-called Raabe criterion, to prove that the following series for any ${\displaystyle s>0}$ :
${\displaystyle \sum _{n=1}^{\infty }{\binom {s}{n}}}$

Solution (Raabe criterion)

Part 1:

• For convergence, we equivalently transform
{\displaystyle {\begin{aligned}\left|{\frac {a_{n+1}}{a_{n}}}\right|\leq 1-{\frac {c}{n+1}}&{\overset {\cdot (n+1)|a_{n}|}{\iff }}(n+1)\left|a_{n+1}\right|\leq (n+1)|a_{n}|-c|a_{n}|\\[0.5em]&\ {\color {Gray}\left\downarrow \ {\text{ add }}(c-1)|a_{n}|{\text{ and subtract }}(n+1)|a_{n+1}|{\text{ on both sides}}\right.}\\[0.5em]&\iff (c-1)\left|a_{n}\right|\leq n|a_{n}|-(n+1)|a_{n+1}|\\[0.5em]&\ {\color {Gray}\left\downarrow \ {\text{ divide both sides by }}c-1>0\right.}\\[0.5em]&\iff \left|a_{n}\right|\leq {\frac {1}{c-1}}\cdot \left(n|a_{n}|-(n+1)|a_{n+1}|\right)\end{aligned}}}

This works for almost all ${\displaystyle n\in \mathbb {N} }$ . So we can find an ${\displaystyle n_{0}\in \mathbb {N} }$, such that the re-formulation works for all ${\displaystyle n\geq n_{0}}$ . We take the sum of both sides starting from ${\displaystyle n_{0}}$ up to some ${\displaystyle N>n_{0}}$:

{\displaystyle {\begin{aligned}\sum _{n=n_{0}}^{N}\left|a_{n}\right|&\leq {\frac {1}{c-1}}\cdot \sum _{n=n_{0}}^{N}\left(n|a_{n}|-(n+1)|a_{n+1}|\right)\\[0.5em]&\ {\color {Gray}\left\downarrow \ {\text{ telescoping sum}}\right.}\\[0.5em]&={\frac {1}{c-1}}\cdot \left(n_{0}|a_{n_{0}}|-(N+1)|a_{N+1}|\right)\\[0.5em]&\leq {\frac {1}{c-1}}\cdot n_{0}|a_{n_{0}}|\end{aligned}}}

Therefore, the sequence of partial sums ${\displaystyle \left(\sum _{n=n_{0}}^{N}\left|a_{n}\right|\right)_{N\in \mathbb {N} }}$ is bounded. That means, the series ${\displaystyle \sum _{n=n_{0}}^{\infty }\left|a_{n}\right|}$ converges absolutely. Adding a finite number of elements, we get that also ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ converges absolutely.

• In the case, where divergence should hold, we can equivalently transform for all ${\displaystyle n\geq n_{0}}$ :
{\displaystyle {\begin{aligned}\left|{\frac {a_{n+1}}{a_{n}}}\right|\geq 1-{\frac {1}{n+1}}&{\overset {\cdot (n+1)|a_{n}|}{\iff }}(n+1)\left|a_{n+1}\right|\geq (n+1)|a_{n}|-|a_{n}|\\[0.5em]&\iff (n+1)\left|a_{n+1}\right|\geq n|a_{n}|\\[0.5em]&\ {\color {Gray}\left\downarrow \ {\text{ subsequently apply the inequality}}\right.}\\[0.5em]&\iff (n+1)\left|a_{n+1}\right|\geq n|a_{n}|\geq (n-1)|a_{n-1}|\geq \ldots \geq n_{0}|a_{n_{0}}|\end{aligned}}}

This is equivalent to

${\displaystyle \left|a_{n+1}\right|\geq n_{0}|a_{n_{0}}|\cdot {\frac {1}{n+1}}}$

The harmonic series ${\displaystyle \sum _{n=n_{0}}^{\infty }\underbrace {n_{0}\left|a_{n_{0}}\right|} _{\text{fest}}\cdot {\frac {1}{n+1}}}$ diverges. By direct comparison, this implies that also ${\displaystyle \sum _{n=n_{0}}^{\infty }a_{n}}$ diverges. Adding a finite number of elements, also ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ diverges, which was to be shown.

Part 2: Here, ${\displaystyle a_{n}={\binom {s}{n}}}$, which means

{\displaystyle {\begin{aligned}\left|{\frac {a_{n+1}}{a_{n}}}\right|&={\frac {\binom {s}{n+1}}{\binom {s}{n}}}\\[0.5em]&=\left|{\dfrac {\frac {s(s-1)\cdot \ldots \cdot (s-n+1)(\overbrace {s-(n+1)+1} ^{=s-n})\cdot n!}{(n+1)!}}{\frac {s(s-1)\cdot \ldots \cdot (s-n+1)}{n!}}}\right|\\[0.5em]&=\left|{\frac {s(s-1)\cdot \ldots \cdot (s-n+1)(s-n)n!}{s(s-1)\cdot \ldots \cdot (s-n+1)\cdot (n+1)n!}}\right|\\[0.5em]&={\frac {|s-n|}{n+1}}\\[0.5em]&\ {\color {Gray}\left\downarrow \ {\text{ for }}n>s\right.}\\[0.5em]&={\frac {n-s}{n+1}}\\[0.5em]&={\frac {n+1-s-1}{n+1}}\\[0.5em]&=1-{\frac {s+1}{n+1}}\\[0.5em]&\leq 1-{\frac {s+1}{n+1}}\end{aligned}}}

So ${\displaystyle c=s+1>1}$ is a suitable constant to use the Raabe criterion and imply absolute convergence of the series ${\displaystyle \sum _{n=1}^{\infty }{\binom {s}{n}}}$.