Proving convergence or divergence of series: Tips and Tricks

Question

I currently write an article (in German) where I collect some tips for students for proving the convergence or divergence of series. What tips and tricks do you know or use or teach?

Remark: I will add some tips as community-wiki answers. So feel free to enhance them.

Answer 1

As a start one should check the term test. When you investigate a series $\displaystyle\sum_{k=1}^\infty a_k$ and $(a_k)_{k\in\mathbb N}$ does not converge to zero, then $\displaystyle\sum_{k=1}^\infty a_k$ diverges.

Answer 2

If the series is of the form $\displaystyle\sum_{k=1}^\infty a_k^k$, the root test might be useful. It states that in case $\displaystyle\lim_{k\to\infty} |a_k| < 1$ the series converges and in the case $\displaystyle\lim_{k\to\infty} |a_k| > 1$ the series diverges.

Example: Take the series $\displaystyle\sum_{k=1}^\infty \left(\frac{4k+5}{2k+3}\right)^k$. For the root test we compute

$$\begin{align} \limsup_{k\to\infty} \sqrt[k]{\left(\frac{4k+5}{2k+3}\right)^k} &= \limsup_{k\to\infty} \frac{4k+5}{2k+3} \\[0.5em] &= \limsup_{k\to\infty} \frac{4+\frac 5k}{2+\frac 3k} \\[0.5em] &= \frac 42 = 2 \end{align}$$

Since $2 > 1$ the series diverges as stated in the root test.

Answer 3

Here is a neat test that is relatively unknown. In any case I can think of where it would be practical to apply, I prefer a direct comparison or limit comparison, but it's certainly still interesting and useful.

Consider the series $\sum_{n=0}^\infty a_n$. Suppose you have a sequence $\{b_n\}_{n=0}^\infty$ such that $\sum_{n=0}^\infty 1/b_n$ diverges. Then let

$$S = \liminf_{n \to \infty} \left(b_n \frac{a_n}{a_{n+1}}-b_{n+1}\right) \ \ \text{ and } \ \ T = \limsup_{n \to \infty} \left(b_n \frac{a_n}{a_{n+1}}-b_{n+1}\right)$$

If $S>0$, then $\sum_{n=0}^\infty a_n$ converges; if $T<0$, then $\sum_{n=0}^\infty a_n$ diverges.

This is known as Kummer's test. It is a generalization of the ratio test (and a number of other tests) which can be made arbitrarily strong: Let

$$c_{0,n}=1, \ \ c_{1,n} = n, \ \ c_{2,n} = n\log{n}, \ \ c_{3,n} = n\log{n}\log\log{n}, \ \ \text{ etc.}$$

Then it is easy to show (e.g. with the integral test) that $\sum_{n=0}^\infty 1/c_{k,n}$ diverges for each $k \in \mathbb{Z}_+$, and also that $\sum_{n=0}^\infty 1/c_{k+1,n}$ diverges slower than $\sum_{n=0}^\infty 1/c_{k,n}$ for each $k \in \mathbb{Z}_+$.

For example, using $c_{0,n}$ gives the ratio test. However, the ratio test is oftentimes inconclusive, e.g. if $a_n = 1/(n(n-1))$. However, if we use $c_{1,n}=n$, we get

$$\begin{align*} S = T & = \lim_{n \to \infty} \left(n \frac{(n+1)n}{n(n-1)}-(n+1)\right) \\ & = \lim_{n \to \infty} \frac{n+1}{n-1}(n-(n-1)) \\ & = 1>0 \\ \end{align*}$$

Therefore, $\sum_{n=0}^\infty a_n$ converges.

Answer 4

For a series of positive terms of the form $$\sum_{n=1}^{\infty}\frac{1}{n^p}$$

a. it converges if $p>1$

b. it diverges if $p\le 1$ (p-series test)

For eg.-$$\sum_{n=1}^{\infty}\frac{2n+3}{n^2+5}$$ diverges as for $n\rightarrow\infty$, $u_n=\frac{2n+3}{n^2+5}\approx\frac{2n}{n^2}\approx\frac{1}{n}$.