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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.

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    The value of this is $\dfrac {\log(2)^3}3$. Proving it is another matter! :-)2012-07-27
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    "A *few* solutionS"? Are you an optimist or what!2012-07-27
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    Now you guys have made me curious. Clearly, this is a known fact to some. I can haz reference, pleez?2012-07-27
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    @HaraldHanche-Olsen: It is an old IMC problem (International Mathematics Competition). The 2012 IMC starts tomorrow morning!2012-07-27
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    @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

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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 $$\frac{\left(\log2\right)^{3}}{3}.$$

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    Great! Now surely you can to this...$$\sum_{n = 1}^{\infty} \operatorname{ln} \biggl(1 + \frac{1}{n}\biggr) \operatorname{ln} \biggl(1 + \frac{1}{2 n}\biggr) = \frac{10 \operatorname{ln} (3)^{5/3}}{49 \operatorname{ln} (2)^{9/4}}$$ Or is that wrong?2012-07-27
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    @GEdgar: sorry but it's wrong (the numerical error is of order $10^{-9}\cdots$) but I subscribe to the principle! :-) So what is this series?2012-07-27
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    @GEdgar: I was not able to get a closed form for your series but could rewrite it as $$\frac {\log(2)^2}2+ \sum_{n=1}^{\infty}\log\left(1+\frac 1{2n}\right)^2$$ with $$\sum_{n=1}^{\infty}\log\left(1+\frac 1{2n}\right)^2=-\sum_{k=1}^\infty \left(-\frac 12\right)^k \frac {\zeta(k+1)H_k}{k+1}$$ this proves : $$\sum_{n=1}^{\infty}\log\left(1+\frac 1{2n}\right)\log\left(1+\frac 1{2n+1}\right)=\frac{\log(2)^2}2$$2012-07-27
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    @Raymond: based on the original sum, this was a joke to show that agreeing to 7 places can happen by "accident".2012-07-27
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    @GEdgar: yes but your little problem is a good one! :-)2012-07-27