# Proving convergence or divergence of series: Tips and Tricks

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.

• Really, the last thing students need is "tips and tricks" for infinite series. Way too many of them already proceed down the rows in a table of results in hopes of finding, by mechanical means, some test that might give the answer. This subject is a wonder of beginning calculus. I would say the vast majority of students come away from it with essentially zero understanding, even if some of the "bright ones" succeed (as measured by exam scores) by mere pattern recognition in the undistinguished sense. – zhw. Jul 20 '16 at 23:49

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.

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 $n\rightarrow\infty$, $u_n=\frac{2n+3}{n^2+5}\approx\frac{2n}{n^2}\approx\frac{1}{n}$.

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.

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.