Question Number 108787 by 150505R last updated on 19/Aug/20
Answered by Dwaipayan Shikari last updated on 19/Aug/20
$$\mathrm{1}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{8}{n}−\mathrm{1}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{8}{n}+\mathrm{1}} \\ $$$$\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{8}{n}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{8}{n}−\mathrm{1}}\right) \\ $$$$\mathrm{1}−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{64}{n}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{8}−\pi−\sqrt{\mathrm{2}}\pi\right)\right) \\ $$$$=\frac{\pi}{\mathrm{8}}+\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{8}} \\ $$$$ \\ $$
Commented by 150505R last updated on 19/Aug/20
$${can}\:{you}\:{explain}\:{fourth}\:{step}\:? \\ $$
Commented by mnjuly1970 last updated on 19/Aug/20
$$\:\:\:\:\:\:\:\:\:\:\:{he}\:{has}\:{used}\:\:{the}\:{digamma} \\ $$$${function}\:. \\ $$
Commented by 1549442205PVT last updated on 19/Aug/20
$$\psi\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{ln}\Gamma\left(\mathrm{x}\right)=\frac{\Gamma'\left(\mathrm{x}\right)}{\Gamma\left(\mathrm{x}\right)},\psi\left(\mathrm{z}+\mathrm{1}\right)=\psi\left(\mathrm{z}\right)+\frac{\mathrm{1}}{\mathrm{z}} \\ $$$$\psi\left(\mathrm{n}\right)=\mathrm{H}_{\mathrm{n}−\mathrm{1}} −\gamma,\mathrm{where}\:\mathrm{H}_{\mathrm{n}} \:=\:\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\Sigma}}\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$\left(\mathrm{H}_{\mathrm{n}} −\mathrm{harmonic}\:\mathrm{number}\right. \\ $$$$\left.,\gamma−\mathrm{Ole}−\mathrm{Mascheroni}\:\mathrm{constant}\right) \\ $$$$\psi\left(\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\right)=−\gamma−\mathrm{2ln2}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\Sigma}}\frac{\mathrm{2}}{\mathrm{2k}−\mathrm{1}} \\ $$