Serlo: EN: Application of convergence criteria

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

If you find out that $(a_{k})_{k\in \mathbb {N} }$ converges to 0, the next question is: How fast does it converge to 0? If it goes to 0 slower than ${\tfrac {1}{k}}$ (harmonic series) you can expect divergence. The harmonic series is the "fastest decaying series that still diverges". By contrast, if $(a_{k})_{k\in \mathbb {N} }$ decays faster than $q^{k}$ for some $q<1$ (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 $(a_{k})_{k\in \mathbb {N} }$ 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 $\sum _{k=1}^{\infty }{\frac {a_{k}}{b_{k}}}$ . If $\limsup _{k\to \infty }\left|{\tfrac {\tfrac {a_{k+1}}{b_{k+1}}}{\tfrac {a_{k}}{b_{k}}}}\right|=\limsup _{k\to \infty }\left|{\tfrac {a_{k+1}b_{k}}{b_{k+1}a_{k}}}\right|<1$, then the series converges absolutely by the ratio test, since it is bounded by some geometric series $\sum _{k=1}^{\infty }c\cdot q^{k}$ (with some constant $c$). In that case, any $\limsup _{k\to \infty }\left|{\tfrac {a_{k+1}b_{k}}{b_{k+1}a_{k}}}\right|<q<1$ serves for such a bound. However, for $\liminf _{k\to \infty }\left|{\tfrac {a_{k+1}b_{k}}{b_{k+1}a_{k}}}\right|>1$, the series diverges. In case $\liminf _{k\to \infty }\left|{\tfrac {a_{k+1}b_{k}}{b_{k+1}a_{k}}}\right|=1=\limsup _{k\to \infty }\left|{\tfrac {a_{k+1}b_{k}}{b_{k+1}a_{k}}}\right|$, 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 $\sum _{k=1}^{\infty }a_{k}^{k}$ or $\sum _{k=1}^{\infty }a_{k}^{k^{2}}$. Absolute convergence holds for $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}^{k}|}}=\limsup _{k\to \infty }|a_{k}|<1$ or $\limsup _{k\to \infty }{\sqrt[{k}]{|a_{k}^{k^{2}}|}}=\limsup _{k\to \infty }|a_{k}^{k}|<1$ . However, for $\limsup _{k\to \infty }|a_{k}|>1$ or $\limsup _{k\to \infty }|a_{k}^{k}|>1$ we have divergence. If the sequence $(a_{k})_{k\in \mathbb {N} }$ or $(a_{k}^{k})_{k\in \mathbb {N} }$ converges, we can replace the $\limsup$ with a $\lim$. If the $\limsup$ equals $1$, 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 $\sum _{k=1}^{\infty }(-1)^{k}a_{k}$ or $\sum _{k=1}^{\infty }(-1)^{k+1}a_{k}$ converges if $(a_{k})_{k\in \mathbb {N} }$ is a monotonously decreasing null sequence. "Monotonously decreasing" makes sure that $a_{2k}-a_{2k+1}=|a_{2k}|-|a_{2k+1}|\geq 0$ (if $a_{2k}$ are the positive elements). By "null sequence", we know that all $a_{2k}-a_{2k+1}$ sum up to $a_{0}$ at most and that the $a_{k}$ are positive. There are alternating series, for which $(a_{k})_{k\in \mathbb {N} }$ does not meet these two criteria. Then, the alternating series test does not work - even though we have an alternating series. If $(a_{k})_{k\in \mathbb {N} }$ is not a null sequence, the series even diverges by the term test. If $(a_{k})_{k\in \mathbb {N} }$ 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 $\sum _{k=1}^{\infty }{\tfrac {P(k)}{Q(k)}}$. The ratio test and the root test may fail in this case. In particular, $P$ and $Q$ might be complicated and therefore difficult to handle. It is easier to compare the fraction to a convergent series $\sum _{k=1}^{\infty }{\tfrac {1}{k^{2}}}={\tfrac {\pi ^{2}}{6}}$ (i.e. power 2 in the denominator) or to a divergent harmonic series $\sum _{k=1}^{\infty }{\tfrac {1}{k}}$ (power 1). In general, convergence can be established by comparison to $\sum _{k=1}^{\infty }{\tfrac {1}{k^{\alpha }}}$ with any power $\alpha >1$ and divergence by comparison to $\sum _{k=1}^{\infty }{\tfrac {1}{k^{\alpha }}}$ with a power $\alpha \leq 1$ . The convergence proof requires bounding $P(k)$ from above and $Q(k)$ from below. The divergence proof works vice versa. If the polynomials $P(k),Q(k)$ have some degrees ${\text{deg}}(P),{\text{deg}}(Q)$, the following rule of thumb holds: If ${\text{deg}}(P)<{\text{deg}}(Q)-1$, the series converges . If ${\text{deg}}(P)\geq {\text{deg}}(Q)-1$ ,it diverges. For a mathematical proof, we can then use a direct comparison to a series $\sum _{k=1}^{\infty }{\tfrac {c}{k^{\alpha }}}$ with some constants $c,\alpha \in \mathbb {R}$.

Decision tree for convergence and divergence

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

Decision tree for convergence and divergence

Applications

Example 1

We consider the series

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

The coefficient sequence is

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

This is a null sequence. We already proved $\lim _{k\to \infty }{\tfrac {k^{a}}{k!}}=0$ for natural numbers $a\in \mathbb {N}$. The squeeze theorem implies $\lim _{k\to \infty }{\tfrac {k^{a}}{k!}}=0$ for all rational $a\in \mathbb {Q}$. 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:

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

In the limit $k\to \infty$, there is

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

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

Example 2

Next, we consider the sequence

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

Again, we have a null sequence of elements:

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

This will get obvious when we write $a_{k}={\tfrac {k^{k}}{2^{k^{2}}}}=\left({\tfrac {k}{2^{k}}}\right)^{k}$ . There is $\lim _{k\to \infty }{\tfrac {k}{2^{k}}}=0$ , so also $\lim _{k\to \infty }a_{k}=\lim _{k\to \infty }\left({\tfrac {k}{2^{k}}}\right)^{k}=0$. 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 $k$, which suggests using the root test. Handling powers of $k$ by the ratio test is tedious, so we try the root test first, i.e. we take the $k$-th root:

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

Since

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

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

Example 3

Now, we investigate the alternating series

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

The corresponding sequence

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

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

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

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

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

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

Example 4

We consider the following quotient of polynomials:

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

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

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

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

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

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

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

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

Example 5

Our last example is another alternating series:

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

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:

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

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