# An infinite series problem

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

Use the integral test: the sum converges if and only if $\lim_{N \to \infty}\left( \int_1^N \frac{1}{n \ln(n+1)}dn \right)$ is finite.

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

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

The integral then becomes $\int_{\ln(2)}^{\ln(N+1)} \! \frac{ n + 1 }{ n t } \, dt$

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

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

So the integral is then \begin{align} \int_{\ln(2)}^{\ln(N+1)} \! \left( \frac{ 1 }{ t } + \frac{ 1 }{ n t } \right) \, dt &= \left[ \ln n \right]_{\ln(2)}^{\ln(N+1)} +\int_{\ln(2)}^{\ln(N+1)} \! \frac{ 1 }{ n t } \, dt \\ & = \ln \left ( \ln( N+1) \right ) – \ln(\ln 2) + \int_{\ln(2)}^{\ln(N+1)} \! \frac{ 1 }{ n t } \, dt \end{align}

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

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