How to find an exact value of the following series $$\sum_{n=2}^{\infty}\frac{1}{2^n}\sum_{k=1}^{n-1}\frac{1}{k(n-k)}$$ I tried to reduce the problem to computing double integral. I also tried some transformations. I tried to use Taylor series for different functions at point $x=1$. Help, please.
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4See https://math.stackexchange.com/q/2333713/42969. Note that $$\sum_{k=1}^{n-1}\frac{1}{k(n-k)} = \frac 2 n \sum_{k=1}^{n-1}\frac{1}{k} = \frac 2n H_{n-1}$$ – Martin R Apr 14 '24 at 18:03
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@MartinR thank you so much buddy! – perenqi Apr 14 '24 at 18:07
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This is the Cauchy product of $\displaystyle\log2=\sum_{n=1}^\infty\frac1{2^n n}$ with itself. Thus the answer is $(\log2)^2$.
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