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

Use the integral test: the sum converges if and only if $$\underset{N\to \mathrm{\infty }}{lim}\left({\int }_1^N\frac{1}{n\ln (n+1)}dn\right)$$ is finite.

Try the substitution $t=\ln (n+1)$.

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

The integral then becomes $${\int }_{\ln (2)}^{\ln (N+1)}\frac{n+1}{nt}dt$$

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 $$\begin{array}{rl}{\int }_{\ln (2)}^{\ln (N+1)}(\frac{1}{t}+\frac{1}{nt})dt& ={[\ln n]}_{\ln (2)}^{\ln (N+1)}+{\int }_{\ln (2)}^{\ln (N+1)}\frac{1}{nt}dt\\ & =\ln (\ln (N+1))–\ln (\ln 2)+{\int }_{\ln (2)}^{\ln (N+1)}\frac{1}{nt}dt\end{array}$$

$\ln (\ln (N+1))–\ln (\ln 2)$ 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 $\ln (\ln (N+1))–\ln (\ln 2)$.

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 $\frac{1}{n\ln (n+1)}>\frac{1}{(n+1)\ln (n+1)}$, which integrates straightforwardly to $\ln (\ln (n+1))$. Much nicer!