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I have tried for a long time on this: $$\sum_{n=1}^{\infty}\ln\left(1+\frac{1}{n}\right)\ln\left(1+\frac{1}{2n}\right)\ln\left(\frac{1}{2n+1}\right)$$

Even I have no idea whether it could be calculate or not

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It is not hard to prove convergence. Would be pessimistic about explicit closed-form expression. – André Nicolas Jan 8 '13 at 4:44
yes, it's easy to prove convergence. here is another tough problem I haven't solved:calculate the sum $\sum_{n\in S}\frac{1}{n}$, where $S=\{n|n\in \mathbb{N} \text{ and } n \text{ has no digits } 9\}$, that is for example $9, 19, 92 \notin S$, $3,11,18\in S$ – Larry Eppes Jan 8 '13 at 5:05
Does that even converge? – user7530 Jan 8 '13 at 5:10
Apparently so: – user7530 Jan 8 '13 at 5:13
This series was a problem from the second day 2011 International Math Competition. See here: – Cody Feb 17 '13 at 13:39

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