# Serlo: EN: Application of convergence criteria

Zur Navigation springen Zur Suche springen

Many problems in real analysis lectures (and also in applications afterwards) involve the investigation whether a certain series converges or diverges. This page offers you a collection of methods how to tackle such problems. We present strategies that experienced mathematicians use to successfully prove or disprove convergence. These strategies are then applied to some practical examples

## Methods for investigating convergence

### Using the term test

A series ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ can never converge, if the corresponding sequence ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ does not converge to 0. It therefore makes sense to first try to find the limit of ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$. If this limit doesn't exist or is not 0, you instantly know that ${\displaystyle \sum _{k=1}^{\infty }a_{k}}$ diverges

If you find out that ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ converges to 0, the next question is: How fast does it converge to 0? If it goes to 0 slower than ${\displaystyle {\tfrac {1}{k}}}$ (harmonic series) you can expect divergence. The harmonic series is the "fastest decaying series that still diverges". By contrast, if ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ decays faster than ${\displaystyle q^{k}}$ for some ${\displaystyle q<1}$ (geometric series) you know that it converges. The geometric series is "one of the slowest decaying series that still converges". It is a useful idea, to compare ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ to a harmonic or geometric series in order to disprove or prove its convergence.

### Ratio test

This test is virtually a comparison to a geometric series. It is often useful for series in quotient form ${\displaystyle \sum _{k=1}^{\infty }{\frac {a_{k}}{b_{k}}}}$ . If ${\displaystyle \limsup _{k\to \infty }\left|{\tfrac {\tfrac {a_{k+1}}{b_{k+1}}}{\tfrac {a_{k}}{b_{k}}}}\right|=\limsup _{k\to \infty }\left|{\tfrac {a_{k+1}b_{k}}{b_{k+1}a_{k}}}\right|<1}$, then the series converges absolutely by the ratio test, since it is bounded by some geometric series ${\displaystyle \sum _{k=1}^{\infty }c\cdot q^{k}}$ (with some constant ${\displaystyle c}$). In that case, any ${\displaystyle \limsup _{k\to \infty }\left|{\tfrac {a_{k+1}b_{k}}{b_{k+1}a_{k}}}\right| serves for such a bound. However, for ${\displaystyle \liminf _{k\to \infty }\left|{\tfrac {a_{k+1}b_{k}}{b_{k+1}a_{k}}}\right|>1}$, the series diverges. In case ${\displaystyle \liminf _{k\to \infty }\left|{\tfrac {a_{k+1}b_{k}}{b_{k+1}a_{k}}}\right|=1=\limsup _{k\to \infty }\left|{\tfrac {a_{k+1}b_{k}}{b_{k+1}a_{k}}}\right|}$, we cannot say anything about the convergence and have to use a different test.

### Root test

This test is also effectively a comparison to a geometric series. It is particularly useful for power series like ${\displaystyle \sum _{k=1}^{\infty }a_{k}^{k}}$ or ${\displaystyle \sum _{k=1}^{\infty }a_{k}^{k^{2}}}$. Absolute convergence holds for ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}^{k}|}}=\limsup _{k\to \infty }|a_{k}|<1}$ or ${\displaystyle \limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}^{k^{2}}|}}=\limsup _{k\to \infty }|a_{k}^{k}|<1}$ . However, for ${\displaystyle \limsup _{k\to \infty }|a_{k}|>1}$ or ${\displaystyle \limsup _{k\to \infty }|a_{k}^{k}|>1}$ we have divergence. If the sequence ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ or ${\displaystyle (a_{k}^{k})_{k\in \mathbb {N} }}$ converges, we can replace the ${\displaystyle \limsup }$ with a ${\displaystyle \lim }$. If the ${\displaystyle \limsup }$ equals ${\displaystyle 1}$, we again cannot make any conclusions and need a different test.

### Alternating series test

Alternating series call for being treated by the alternating series criterion. According to it, any series ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k}a_{k}}$ or ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k+1}a_{k}}$ converges if ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ is a monotonously decreasing null sequence. "Monotonously decreasing" makes sure that ${\displaystyle a_{2k}-a_{2k+1}=|a_{2k}|-|a_{2k+1}|\geq 0}$ (if ${\displaystyle a_{2k}}$ are the positive elements). By "null sequence", we know that all ${\displaystyle a_{2k}-a_{2k+1}}$ sum up to ${\displaystyle a_{0}}$ at most and that the ${\displaystyle a_{k}}$ are positive. There are alternating series, for which ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ does not meet these two criteria. Then, the alternating series test does not work - even though we have an alternating series. If ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ is not a null sequence, the series even diverges by the term test. If ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ is a null sequence that does not decrease monotonously, we need to search for a different criterion.

### Direct comparison

Direct comparison is useful, if we have a fraction of polynomials - like ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {P(k)}{Q(k)}}}$. The ratio test and the root test may fail in this case. In particular, ${\displaystyle P}$ and ${\displaystyle Q}$ might be complicated and therefore difficult to handle. It is easier to compare the fraction to a convergent series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}={\tfrac {\pi ^{2}}{6}}}$ (i.e. power 2 in the denominator) or to a divergent harmonic series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{k}}}$ (power 1). In general, convergence can be established by comparison to ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{k^{\alpha }}}}$ with any power ${\displaystyle \alpha >1}$ and divergence by comparison to ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{k^{\alpha }}}}$ with a power ${\displaystyle \alpha \leq 1}$ . The convergence proof requires bounding ${\displaystyle P(k)}$ from above and ${\displaystyle Q(k)}$ from below. The divergence proof works vice versa. If the polynomials ${\displaystyle P(k),Q(k)}$ have some degrees ${\displaystyle {\text{deg}}(P),{\text{deg}}(Q)}$, the following rule of thumb holds: If ${\displaystyle {\text{deg}}(P)<{\text{deg}}(Q)-1}$, the series converges . If ${\displaystyle {\text{deg}}(P)\geq {\text{deg}}(Q)-1}$ ,it diverges. For a mathematical proof, we can then use a direct comparison to a series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {c}{k^{\alpha }}}}$ with some constants ${\displaystyle c,\alpha \in \mathbb {R} }$.

## Decision tree for convergence and divergence

The tricks above can be visually represented in a decision tree:

## Applications

### Example 1

We consider the series

${\displaystyle \sum _{k=1}^{\infty }{\frac {k^{a}}{k!}}{\text{ for }}a\in \mathbb {Q} }$

The coefficient sequence is

${\displaystyle (a_{k})_{k\in \mathbb {N} }=\left({\frac {k^{a}}{k!}}\right)_{k\in \mathbb {N} }}$

This is a null sequence. We already proved ${\displaystyle \lim _{k\to \infty }{\tfrac {k^{a}}{k!}}=0}$ for natural numbers ${\displaystyle a\in \mathbb {N} }$. The squeeze theorem implies ${\displaystyle \lim _{k\to \infty }{\tfrac {k^{a}}{k!}}=0}$ for all rational ${\displaystyle a\in \mathbb {Q} }$. The sequence is not alternating, since its elements are positive. As it is a sequence of quotients, we might have good luck with the ratio test:

{\displaystyle {\begin{aligned}\left|{\frac {a_{k+1}}{a_{k}}}\right|&={\frac {\frac {(k+1)^{a}}{(k+1)!}}{\frac {k^{a}}{k!}}}\\[0.5em]&={\frac {(k+1)^{a}k!}{k^{a}(k+1)!}}\\[0.5em]&={\frac {(k+1)^{a}}{k^{a}}}\cdot {\frac {k!}{(k+1)!}}\\[0.5em]&=\left(1+{\frac {1}{k}}\right)^{a}\cdot {\frac {1}{k+1}}\end{aligned}}}

In the limit ${\displaystyle k\to \infty }$, there is

{\displaystyle {\begin{aligned}\lim _{k\to \infty }\left|{\frac {a_{k+1}}{a_{k}}}\right|&=\lim _{k\to \infty }\left(1+{\frac {1}{k}}\right)^{a}\cdot {\frac {1}{k+1}}\\[0.5em]&=1^{a}\cdot 0\\[0.5em]&=0<1\end{aligned}}}

So the ratio test applies and our series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {k^{a}}{k!}}}$ converges absolutely. Done with it!

### Example 2

Next, we consider the sequence

${\displaystyle \sum _{k=1}^{\infty }{\frac {k^{k}}{2^{k^{2}}}}}$

Again, we have a null sequence of elements:

${\displaystyle (a_{k})_{k\in \mathbb {N} }=\left({\frac {k^{k}}{2^{k^{2}}}}\right)_{k\in \mathbb {N} }}$

This will get obvious when we write ${\displaystyle a_{k}={\tfrac {k^{k}}{2^{k^{2}}}}=\left({\tfrac {k}{2^{k}}}\right)^{k}}$ . There is ${\displaystyle \lim _{k\to \infty }{\tfrac {k}{2^{k}}}=0}$ , so also ${\displaystyle \lim _{k\to \infty }a_{k}=\lim _{k\to \infty }\left({\tfrac {k}{2^{k}}}\right)^{k}=0}$. This sequence is again not alternating as all elements are positive. We have a quotient, which suggests taking the ratio test. But there is also a power of ${\displaystyle k}$, which suggests using the root test. Handling powers of ${\displaystyle k}$ by the ratio test is tedious, so we try the root test first, i.e. we take the ${\displaystyle k}$-th root:

{\displaystyle {\begin{aligned}{\sqrt[{k}]{|a_{k}|}}&={\sqrt[{k}]{\frac {k^{k}}{2^{k^{2}}}}}\\[0.5em]&={\sqrt[{k}]{\left({\frac {k}{2^{k}}}\right)^{k}}}\\[0.5em]&={\frac {k}{2^{k}}}\end{aligned}}}

Since

${\displaystyle \lim _{k\to \infty }{\frac {k}{2^{k}}}=0<1}$

the root test succeeds and our series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {k^{k}}{2^{k^{2}}}}}$ converges absolutely.

### Example 3

Now, we investigate the alternating series

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

The corresponding sequence

${\displaystyle (a_{k})_{k\in \mathbb {N} }=\left({\frac {(-1)^{k}}{2k-1}}\right)_{k\in \mathbb {N} }}$

is a null sequence, since ${\displaystyle \lim _{k\to \infty }|a_{k}|=\lim _{k\to \infty }{\tfrac {1}{2k-1}}=0}$. At this point, the alternating series test seems the first option. In order to apply it, we need to check whether ${\displaystyle (b_{k})_{k\in \mathbb {N} }}$ is monotonously decreasing. For all ${\displaystyle k\in \mathbb {N} }$, there is:

{\displaystyle {\begin{aligned}&2k+1\geq 2k-1\\[0.5em]\iff &\underbrace {\frac {1}{2k+1}} _{=a_{k+1}}\leq \underbrace {\frac {1}{2k-1}} _{=a_{k}}\end{aligned}}}

So ${\displaystyle \left({\tfrac {1}{2k-1}}\right)_{k\in \mathbb {N} }}$ is monotonously decreasing. By the alternating series test, we know that the series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {(-1)^{k}}{2k-1}}}$ converges. But does it also converge absolutely? We need to investigate whether the series ${\displaystyle \sum _{k=1}^{\infty }\left|{\tfrac {(-1)^{k}}{2k-1}}\right|=\sum _{k=1}^{\infty }{\tfrac {1}{2k-1}}}$ converges. This scales like a harmonic series, so it should not converge. And indeed, we can compare it to a harmonic series:

${\displaystyle 2k-1\leq 2k\iff {\frac {1}{2k-1}}\geq {\frac {1}{2k}}}$

As the harmonic series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{k}}}$ diverges, we also have divergence of the scaled version ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{2k}}=\sum _{k=1}^{\infty }{\tfrac {1}{2}}\cdot {\tfrac {1}{k}}}$ and by direct comparison, the series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{2k-1}}}$ diverges. So our series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {(-1)^{k}}{2k-1}}}$ is convergent, but not absolutely convergent.

### Example 4

We consider the following quotient of polynomials:

${\displaystyle \sum _{k=1}^{\infty }{\frac {k^{2}-2k+1}{2k^{4}+6}}}$

The coefficient sequence is ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ with

${\displaystyle a_{k}={\frac {k^{2}-2k+1}{2k^{4}+6}}}$

The two polynomials in the fraction ${\displaystyle {\tfrac {P(k)}{Q(k)}}}$ are ${\displaystyle P(k)=k^{2}-2k+1}$ and ${\displaystyle Q(k)=2k^{4}+6}$. Its degrees are

${\displaystyle \underbrace {{\text{deg}}(P)} _{=2}<\underbrace {{\text{deg}}(Q)-1} _{4-1=3}}$

which implies absolute convergence by direct comparison: The coefficient sequence scales like ${\displaystyle {\tfrac {k^{2}}{k^{4}}}={\tfrac {1}{k^{2}}}}$ for large ${\displaystyle k}$. We can therefore bound it from above by ${\displaystyle C\cdot {\tfrac {1}{k^{2}}}}$ with ${\displaystyle C\in \mathbb {R} ^{+}}$ . Explicitly, we can do this by increasing the enumerator and decreasing the denominator:

{\displaystyle {\begin{aligned}|a_{k}|&={\frac {k^{2}-2k+1}{2k^{4}+6}}\\[0.5em]&={\frac {(k-1)^{2}}{2k^{4}+6}}\\[0.5em]&\leq {\frac {k^{2}}{2k^{4}+6}}\\[0.5em]&\leq {\frac {k^{2}}{2k^{4}}}\\[0.5em]&={\frac {1}{2}}{\frac {1}{k^{2}}}\end{aligned}}}

The series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {1}{2}}{\tfrac {1}{k^{2}}}={\tfrac {1}{2}}\sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}={\tfrac {1}{2}}{\tfrac {\pi ^{2}}{6}}={\tfrac {\pi ^{2}}{12}}}$ converges, so also our series ${\displaystyle \sum _{k=1}^{\infty }{\tfrac {k^{2}-2k+1}{2k^{4}+6}}}$ converges absolutely (and ${\displaystyle {\tfrac {\pi ^{2}}{12}}}$ is an upper bound for it).

### Example 5

Our last example is another alternating series:

${\displaystyle \sum _{k=1}^{\infty }(-1)^{k}\left({\frac {k}{k+1}}\right)}$

It may be tempting to use the alternating series test, here. However, we should first check whether the sequence of elements is even a null sequence:

${\displaystyle |a_{k}|={\frac {k}{k+1}}={\frac {1}{1+{\frac {1}{k}}}}{\overset {k\to \infty }{\to }}{\frac {1}{1+0}}=1}$

Apparently, ${\displaystyle (|a_{k}|)_{k\in \mathbb {N} }}$ is not a null sequence! So we can not apply the alternating series test. However, we instantly know that ${\displaystyle (a_{k})_{k\in \mathbb {N} }}$ is neither a null sequence and can directly apply the term test. By means of the term test, our series ${\displaystyle \sum _{k=1}^{\infty }(-1)^{k}\left({\tfrac {k}{k+1}}\right)}$ diverges.