I encountered the following double sum which requires an evaluation.

Is there a closed form for this? $$\sum_{n=0}^{\infty}\frac{\sum_{k=0}^n\binom{n}k^{-1}}{(n+1)(n+2)}.$$ Incidentally, it seems that the following gives (empirically) the same value. Does it? $$\sum_{n=1}^{\infty}\frac1{2^{n-1}(n+1)}\sum_{k=1}^n\frac{\binom{n}{2k-1}}{2k-1}.$$

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    $\begingroup$ looks like $\pi^2/6$. Represent the inverse binomial as a Beta integral and transform the sum under integrand as a geometric progression. $\endgroup$ – Fedor Petrov Jun 3 '18 at 16:18
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    $\begingroup$ $\sum\binom{n}k^{-1}<3$, so the series converges. $\endgroup$ – T. Amdeberhan Jun 3 '18 at 16:20
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    $\begingroup$ 1st time I tried Wolfram Alpha. Maybe last time, ha ha. $\endgroup$ – Mark L. Stone Jun 3 '18 at 16:34
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    $\begingroup$ @MarkL.Stone: Strangely, WA is very sensitive to syntax: in correct Mathematica syntax I get the opposite: "By the comparison test, the series converges." $\endgroup$ – Alex M. Jun 3 '18 at 17:24
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    $\begingroup$ Numerical calculations do not confirm your suggestion, producing for the second sum from 1 to 2000 the value 1.12068617279184 (the code on demand). $\endgroup$ – user64494 Jun 3 '18 at 19:12

Consider the integral $$I=\int_0^1\int_0^1\frac{zdzdt}{(1-zt)(1-z(1-t))}=\sum_{k,j\geqslant 0} \int_0^1\int_0^1 z^{k+j+1}t^k(1-t)^jdzdt=\\ \sum_{k,j\geqslant 0} \frac1{k+j+2}\cdot \frac{k!j!}{(k+j+1)!},$$ that is your sum (denote $n=k+j$). For evaluating the integral, we first integrate by $t$, get $-2\log(1-z)/(2-z)$. Now denote $1-z=x$ and integrate $2\log x/(1+x)=2\log x(1-x+x^2-x^3+\dots)$ using $\int_0^1 -\log x\cdot x^m dx=\frac1{(m+1)^2}$. We get $2(1-1/4+1/9-1/16+\dots)=\pi^2/6$.

As for your second sum, it has indeed the same value which may be obtained similarly. We consider the integral $$ \int_0^1\int_0^1 \frac1x\left(\frac1{1-y\frac{1+x}2}-\frac1{1-y\frac{1-x}2}\right)dxdy=\\ \sum_n \int_0^1y^n dy\cdot 2^{1-n}\int_0^1\frac{(1+x)^n-(1-x)^n}{2x}dx, $$ that is your sum (expand $(1+x)^n-(1-x)^n=2\sum \binom{n}{2k-1}x^{2k-1}$ and integrate).

If we integrate first in $x$, we get $-2\log(1-y)/(2-y)$, the same thing as in the first sum.

  • $\begingroup$ Good job, Fedor. Hope one can also solve (answer) the second part of the problem which I posted after your reply. $\endgroup$ – T. Amdeberhan Jun 4 '18 at 17:49
  • $\begingroup$ @T.Amdeberhan added. I wonder why computations show another answer. $\endgroup$ – Fedor Petrov Jun 5 '18 at 8:42
  • $\begingroup$ It might be that the 2nd series is slowly converging. $\endgroup$ – T. Amdeberhan Jun 5 '18 at 12:57
  • $\begingroup$ It converges as $\sum 1/n^2$, 1000 terms should give the accuracy about $1/1000$. $\endgroup$ – Fedor Petrov Jun 6 '18 at 12:23

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