The question here requires the evaluation of the following summation:

$$\sum_{n=1}^\infty\frac 1{(2n+1)(2n+2)}\sum_{r=1}^n \frac 1r\tag{1}$$

The answers provided (and also wolframalpha) show that the solution is $$\frac {\pi^2}{12}-(\ln 2)^2\tag{2}$$

From the standard expansion of $\ln (1+x)$ and result of the Basel problem, $(2)$ can be written as $$\sum_{n=1}^\infty \frac 1{2n^2}-\left(\sum_{n=1}^\infty \frac {(-1)^{n+1}}n\right)^2\tag{3}$$ and from answers to my other question on the Cauchy Product here, the second term above can be written as $$\left(\sum_{n=1}^\infty \frac {(-1)^{n+1}}n\right)^2=\sum_{n=1}^\infty\sum_{j=1}^n\frac {(-1)^{j+1}}j\cdot \frac {(-1)^{n+2-j}}{n+1-j} =\sum_{n=1}^\infty\sum_{j=1}^n\frac {(-1)^{n+1}}{j(n+1-j)}$$ Hence $(3)$ can be written as $$\sum_{n=1}^\infty \frac 1{2n^2}-\sum_{n=1}^\infty\sum_{j=1}^n\frac {(-1)^{n+1}}{j(n+1-j)}=\sum_{n=1}^\infty \left(\frac 1{2n^2}-\sum_{j=1}^n\frac {(-1)^{n+1}}{j(n+1-j)}\right)\tag{4}$$

Is it possible to derive $(3)$ or $(4)$ directly from $(1)$ without first working out the closed-form result given in $(2)$?

  • $\begingroup$ Is this a follow-up to the (closed) question here - math.stackexchange.com/a/2481928/44121 ? $\endgroup$ – Jack D'Aurizio Oct 24 '17 at 20:12
  • $\begingroup$ @JackD'Aurizio - It is based on that question (the same reference also given in the question above) but in a way this is a different question, because it is asking if the original summation can be expressed directly as the sum (or rather difference) of two summations given. $\endgroup$ – hypergeometric Oct 25 '17 at 14:51

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