# Math for Non-Geeks/ 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 monotonically 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 monotonically, 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:

## 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 monotonically 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 monotonically 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.