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Question Number 108787 by 150505R last updated on 19/Aug/20

Answered by Dwaipayan Shikari last updated on 19/Aug/20

$$1-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{8n-1}+\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{8n+1}$$$$1+\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{1}{8n+1}-\frac{1}{8n-1})$$$$1-2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{64n^2-1}$$$$1-2\left(\frac{1}{16}(8-\pi -\sqrt{2}\pi )\right)$$$$=\frac{\pi }{8}+\frac{\sqrt{2}\pi }{8}$$$$$$

Commented by 150505R last updated on 19/Aug/20

$$canyouexplainfourthstep?$$

Commented by mnjuly1970 last updated on 19/Aug/20

$$hehasusedthedigamma$$$$function.$$

Commented by 1549442205PVT last updated on 19/Aug/20

$$\psi \left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\ln \mathrm{\Gamma }\left(\mathrm{x}\right)=\frac{{\mathrm{\Gamma }}^{\prime }\left(\mathrm{x}\right)}{\mathrm{\Gamma }\left(\mathrm{x}\right)},\psi (\mathrm{z}+1)=\psi \left(\mathrm{z}\right)+\frac{1}{\mathrm{z}}$$$$\psi \left(\mathrm{n}\right)={\mathrm{H}}_{\mathrm{n}-1}-\gamma ,\mathrm{where}{\mathrm{H}}_{\mathrm{n}}=\underset{\mathrm{k}=1}{\stackrel{\mathrm{n}}{\mathrm{\Sigma }}}\frac{1}{\mathrm{k}}$$$$({\mathrm{H}}_{\mathrm{n}}-\mathrm{harmonic}\mathrm{number}$$$$,\gamma -\mathrm{Ole}-\mathrm{Mascheroni}\mathrm{constant})$$$$\psi (\mathrm{n}+\frac{1}{2})=-\gamma -2\ln 2+\underset{\mathrm{k}=1}{\stackrel{\mathrm{n}}{\mathrm{\Sigma }}}\frac{2}{2\mathrm{k}-1}$$

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