Exercises: Convergence criteria for series – Serlo

Application of convergence criteria

Exercise (Convergence proof training 1)

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

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

Solution (Convergence proof training 1)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence, the series converges absolutely.

6. Alternating series test:

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

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

So

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

So the series converges.

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

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

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

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

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

Note: The $(-1)^{k}$ 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, $(b_{k})=({\tfrac {1}{\sqrt[{k}]{k}}})$ is not a null sequence, which is why the alternating series test doesn't work.

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

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

Therefore, the series converges.

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

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

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

Exercise (Convergence proof training 2)

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

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

Solution (Convergence proof training 2)

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

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

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

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

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

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

So the series diverges.

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

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

So the series converges absolutely.

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

Alternative: Root est:

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

So the series converges absolutely.

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

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

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

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

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

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

By the alternating series test, the series converges.

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

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

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

Therefore, the series converges absolutely.

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

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

So the series converges absolutely.

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

Exercise (Series depending on parameters)

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

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

Solution (Series depending on parameters)

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

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

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

2. Case distinction:

Fall 1: $|x|\leq 1$

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

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

Fall 2: $|x|>1$

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

So the series diverges.

3. Case distinction:

Wir unterscheiden zwei Fälle:

Fall 1: $|x|<1$

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

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

So indeed, the series diverges absolutely.

Fall 2: $|x|\geq 1$

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

Therefore, the series diverges.

4. Case distinction:

Fall 1: $|x|<1$

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

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

Fall 2: $x=1$

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

Fall 3: $x=-1$

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

Fall 4: $|x|>1$

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

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

Exercise (Direct comparison in asymptotic behaviour)

Investigate, whether the series

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

converges or diverges.

Solution (Direct comparison in asymptotic behaviour)

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

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

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

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

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

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

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

Term test - strengthened version

Exercise (Term test - strengthened version)

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

Solution (Term test - strengthened version)

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

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

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

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

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

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

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

Now we are still missing the odd numbers:

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

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

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

Hence, for all $k\geq N$:

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

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

The case distinction was actually only necessary since for even numbers $K=2k$, we had to sum over $k={\tfrac {K}{2}}$ and for odd numbers $K=2k-1$ over $k={\tfrac {K+1}{2}}$ elements. Since both for even $(2ka_{2k})$ and odd indices $((2k-1)a_{2k-1})$ , the elements tend to zero as $k\to \infty$, the sequence $(na_{n})$ 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 $\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {1}{k}}$ converges. (note: the harmonic series $\sum _{k=1}^{\infty }{\frac {1}{k}}$ does not converge!)

Solution (Alternating harmonic series)

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

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

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

Root and ratio test: estimate of errors

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

Exercise (Error estimate for the root test)

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

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

Solution (Error estimate for the root test)

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

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

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

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

Exercise (Error estimate for the ratio test)

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

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

Solution (Error estimate for the ratio test)

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

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

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

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

Again, we sum up the geometric series bounds:

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

Exercise (showing that a sequence is null)

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

Solution (showing that a sequence is null)

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

${\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }{\frac {{\binom {n+1}{k}}q^{n+1}}{{\binom {n}{k}}q^{n}}}\\[0.3em]&=\lim _{n\to \infty }{\frac {{\frac {(n+1)n(n-1)\cdot \ldots \cdot ((n+1)-k+1)}{k!}}\cdot q\cdot q^{n}}{{\frac {n(n-1)\cdot \ldots \cdot (n-k+1)}{k!}}\cdot q^{n}}}\\[0.3em]&=\lim _{n\to \infty }{\frac {(n+1)n(n-1)\cdot \ldots \cdot (n-k+2)}{n(n-1)\cdot \ldots \cdot (n-k+2)(n-k+1)}}\cdot q\\[0.3em]&=\lim _{n\to \infty }{\frac {n+1}{n-k+1}}\cdot q\\[0.3em]&=\lim _{n\to \infty }{\frac {1+\overbrace {\frac {1}{n}} ^{\to 0}}{1-\underbrace {\frac {k}{n}} _{\to 0}+\underbrace {\frac {1}{n}} _{\to 0}}}\cdot q\\[0.3em]&=q<1\end{aligned}}$

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

Alternating series test with error bounds

Exercise (Alternating series test with error bounds)

First, show that the series

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

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

Solution (Alternating series test with error bounds)

Proof step: The series converges

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

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

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

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

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

Proof step: Finding $n_{0}$

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

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

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

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

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

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

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

Cauchy condensation test

Exercise (A double logarithm)

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

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

Solution (A double logarithm)

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

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

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

Fall 1: $\alpha >1$

Here,

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

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

Fall 2: $0<\alpha \leq 1$

Here,

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

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

Further convergence criteria

Exercise (Series with products)

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

1. Whenever the series $\sum _{k=1}^{\infty }a_{k}$ converges absolutely and $(b_{k})_{k\in \mathbb {N} }$ is bounded, then also $\sum _{k=1}^{\infty }a_{k}b_{k}$ converges absolutely.
2. There are pairs of a series $\sum _{k=1}^{\infty }a_{k}$ , which converges and a sequence $(b_{k})_{k\in \mathbb {N} }$ which is bounded, such that $\sum _{k=1}^{\infty }a_{k}b_{k}$ 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 $\sum _{k=1}^{\infty }a_{k}$ converges absolutely, the sequence of partial sums $\left(\sum _{k=1}^{n}|a_{k}|\right)_{n\in \mathbb {N} }$ must be bounded. Further, $(b_{k})_{k\in \mathbb {N} }$ is bounded. That means, there is an upper bound $S>0$ with $|b_{k}|\leq S$ for all $k\in \mathbb {N}$ and therefore

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

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

Way 2: Direct comparison test

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

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

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

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

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

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

Note: other counter-examples are also possible.

Exercise (Raabe criterion)

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

Solution (Raabe criterion)

Part 1:

• For convergence, we equivalently transform

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

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

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

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

• In the case, where divergence should hold, we can equivalently transform for all $n\geq n_{0}$ :

${\begin{aligned}\left|{\frac {a_{n+1}}{a_{n}}}\right|\geq 1-{\frac {1}{n+1}}&{\overset {\cdot (n+1)|a_{n}|}{\iff }}(n+1)\left|a_{n+1}\right|\geq (n+1)|a_{n}|-|a_{n}|\\[0.5em]&\iff (n+1)\left|a_{n+1}\right|\geq n|a_{n}|\\[0.5em]&\ {\color {Gray}\left\downarrow \ {\text{ subsequently apply the inequality}}\right.}\\[0.5em]&\iff (n+1)\left|a_{n+1}\right|\geq n|a_{n}|\geq (n-1)|a_{n-1}|\geq \ldots \geq n_{0}|a_{n_{0}}|\end{aligned}}$

This is equivalent to

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

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

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

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

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