4
$\begingroup$

I was told this interesting question today, but I haven't managed to get very far:

Evaluate $\sum_{n=1}^\infty \log \left(1+\frac{1}{n}\right)\log \left(1+\frac{1}{2n}\right)\log \left(1+\frac{1}{2n+1}\right).$

I am interested in seeing at least a few solutions.

  • 0
    @EricNaslund: I should have guessed it's something of that sort. It certainly has that peculiar competition flavour to it …2012-07-27

1 Answers 1

7

Here is a solution I just found. Notice that $\log\left(1+\frac{1}{2n+1}\right)=\log\left(1+\frac{1}{n}\right)-\log\left(1+\frac{1}{2n}\right)$ so that our series becomes $\sum_{n=1}^{\infty}\left(\log\left(1+\frac{1}{n}\right)^{2}\log\left(1+\frac{1}{2n}\right)-\log\left(1+\frac{1}{n}\right)\log\left(1+\frac{1}{2n}\right)^{2}\right).$ Since $\log\left(1+\frac{1}{2n+1}\right)^{3}=\log\left(1+\frac{1}{n}\right)^{3}-3\log\left(1+\frac{1}{n}\right)^{2}\log\left(1+\frac{1}{2n}\right)+3\log\left(1+\frac{1}{n}\right)\log\left(1+\frac{1}{2n}\right)^{2}-\log\left(1+\frac{1}{2n}\right)^{3},$ we see that our series equals $\frac{1}{3}\left(\sum_{n=1}^{\infty}\log\left(1+\frac{1}{n}\right)^{3}-\log\left(1+\frac{1}{2n}\right)^{3}-\log\left(1+\frac{1}{2n+1}\right)^{3}\right),$ and the above telescopes and equals $\frac{\left(\log2\right)^{3}}{3}.$

  • 0
    @GEdgar: yes but your little problem is a good one! :-)2012-07-27