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!