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Question Number 116997 by mathdave last updated on 08/Oct/20

Answered by Olaf last updated on 09/Oct/20

$$\frac{1}{(2n+1)(2n+2+k)}=\frac{1}{k+1}[\frac{1}{2n+1}-\frac{1}{2n+2+k}]$$$$\mathrm{S}=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k}{k+1}\underset{n=0}{\sum ^{\mathrm{\infty }}}[\frac{{\mathrm{C}}_{2n}^n}{2^{2n}(2n+1)}-\frac{{\mathrm{C}}_{2n}^n}{2^{2n}(2n+2+k)}]$$$$\mathrm{S}=\underset{n=0}{\sum ^{\mathrm{\infty }}}[\frac{{\mathrm{C}}_{2n}^n}{2^{2n}(2n+1)}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k}{k+1}-\frac{{\mathrm{C}}_{2n}^n}{2^{2n}}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k}{2n+2+k}]$$$$\mathrm{S}=\underset{n=0}{\sum ^{\mathrm{\infty }}}[-\frac{{\mathrm{C}}_{2n}^n}{2^{2n}(2n+1)}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{k+1}}{k+1}-\frac{{\mathrm{C}}_{2n}^n}{2^{2n}}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{2n+2+k}}{2n+2+k}]$$$$\mathrm{S}=\underset{n=0}{\sum ^{\mathrm{\infty }}}[-\frac{{\mathrm{C}}_{2n}^n}{2^{2n}(2n+1)}\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k}{k}-\frac{{\mathrm{C}}_{2n}^n}{2^{2n}}\underset{k=2n+2}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k}{k}]$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k}{k}=-\ln 2$$$$\mathrm{S}=\underset{n=0}{\sum ^{\mathrm{\infty }}}[\ln 2\frac{{\mathrm{C}}_{2n}^n}{2^{2n}(2n+1)}-\frac{{\mathrm{C}}_{2n}^n}{2^{2n}}(-\ln 2-\underset{k=1}{\sum ^{k=2n+1}}\frac{{(-1)}^k}{k})]$$$$\mathrm{S}=\ln 2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{C}}_{2n}^n}{2^{2n}}(\frac{1}{2n+1}+1)+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{C}}_{2n}^n}{2^{2n}}\underset{k=1}{\sum ^{k=2n+1}}\frac{{(-1)}^k}{k}$$$$\mathrm{S}=\ln 2\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{C}}_{2n}^n}{2^{2n}}.\frac{2n+2}{2n+1}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{C}}_{2n}^n}{2^{2n}}\underset{k=1}{\sum ^{k=2n+1}}\frac{{(-1)}^k}{k}$$$$workinprogress...$$

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