# An infinite series problem

A couple of undergrads came into maths-aid today with a problem: does the series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\mathrm{ln}\left(n+1\right)}$ converge?

Use the integral test: the sum converges if and only if $\underset{N\to \mathrm{\infty }}{lim}\left({\int }_{1}^{N}\frac{1}{n\mathrm{ln}\left(n+1\right)}dn\right)$ is finite.

Try the substitution $t=\mathrm{ln}\left(n+1\right)$.

$\frac{dt}{dn}=\frac{1}{n+1}$, so $dn=\left(n+1\right)dt$.

The integral then becomes ${\int }_{\mathrm{ln}\left(2\right)}^{\mathrm{ln}\left(N+1\right)}\phantom{\rule{-0.167em}{0ex}}\frac{n+1}{nt}\phantom{\rule{0.167em}{0ex}}dt$

I’ve still got $n$ in there a couple of times, but bear with me…

You can split the fraction into two bits: $\frac{n+1}{nt}=\frac{n}{nt}+\frac{1}{nt}=\frac{1}{t}+\frac{1}{nt}$

So the integral is then $\begin{array}{rl}{\int }_{\mathrm{ln}\left(2\right)}^{\mathrm{ln}\left(N+1\right)}\phantom{\rule{-0.167em}{0ex}}\left(\frac{1}{t}+\frac{1}{nt}\right)\phantom{\rule{0.167em}{0ex}}dt& ={\left[\mathrm{ln}n\right]}_{\mathrm{ln}\left(2\right)}^{\mathrm{ln}\left(N+1\right)}+{\int }_{\mathrm{ln}\left(2\right)}^{\mathrm{ln}\left(N+1\right)}\phantom{\rule{-0.167em}{0ex}}\frac{1}{nt}\phantom{\rule{0.167em}{0ex}}dt\\ & =\mathrm{ln}\left(\mathrm{ln}\left(N+1\right)\right)–\mathrm{ln}\left(\mathrm{ln}2\right)+{\int }_{\mathrm{ln}\left(2\right)}^{\mathrm{ln}\left(N+1\right)}\phantom{\rule{-0.167em}{0ex}}\frac{1}{nt}\phantom{\rule{0.167em}{0ex}}dt\end{array}$

$\mathrm{ln}\left(\mathrm{ln}\left(N+1\right)\right)–\mathrm{ln}\left(\mathrm{ln}2\right)$ is unbounded as $N$ tends to infinity, but what about the other integral, that I can’t see how to do?

It doesn’t matter, because it’s definitely positive. So I’m sure the total integral is at least $\mathrm{ln}\left(\mathrm{ln}\left(N+1\right)\right)–\mathrm{ln}\left(\mathrm{ln}2\right)$.

The integral is unbounded, hence the series diverges.

(I wrote this post using my write maths, see maths thing. It was very easy!)

Update: A couple of other Newcastle PhDs, David Cushing and David Elliott, both pointed out that $\frac{1}{n\mathrm{ln}\left(n+1\right)}>\frac{1}{\left(n+1\right)\mathrm{ln}\left(n+1\right)}$, which integrates straightforwardly to $\mathrm{ln}\left(\mathrm{ln}\left(n+1\right)\right)$. Much nicer!