Question Number 128236 by Dwaipayan Shikari last updated on 05/Jan/21
$$\frac{\mathrm{1}}{\mathrm{6}}\underset{{p}} {\overset{\infty} {\sum}}\frac{{logp}}{{p}^{\mathrm{2}} −\mathrm{1}}=\frac{{log}\mathrm{1}}{\pi^{\mathrm{2}} }+\frac{{log}\mathrm{2}}{\mathrm{4}\pi^{\mathrm{2}} }+\frac{{log}\mathrm{3}}{\mathrm{9}\pi^{\mathrm{2}} }+…\:\:\:\left({p}={prime}\right) \\ $$
Commented by Dwaipayan Shikari last updated on 05/Jan/21
$$\zeta\left({s}\right)=\underset{{p}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}^{{s}} }\right)^{−\mathrm{1}} \\ $$$${log}\left(\zeta\left({s}\right)\right)=−\underset{{p}} {\overset{\infty} {\sum}}{log}\left(\mathrm{1}−\frac{\mathrm{1}}{{p}^{{s}} }\right) \\ $$$$\frac{\zeta'\left({s}\right)}{\zeta\left({s}\right)}=−\overset{\infty} {\sum}\frac{{p}^{−{s}} {log}\left({p}\right)}{\mathrm{1}−\frac{\mathrm{1}}{{p}^{{s}} }}\Rightarrow\zeta'\left({s}\right)=\zeta\left({s}\right)\underset{{p}} {\overset{\infty} {\sum}}\frac{{logp}}{\mathrm{1}−{p}^{{s}} } \\ $$$$\zeta'\left({s}\right)=−\overset{\infty} {\sum}\frac{{logn}}{{n}^{{s}} } \\ $$$$\Rightarrow\frac{{log}\mathrm{1}}{\mathrm{1}}+\frac{{log}\mathrm{2}}{\mathrm{2}^{\mathrm{2}} }+\frac{{log}\mathrm{3}}{\mathrm{3}^{\mathrm{2}} }+…=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\underset{{p}} {\overset{\infty} {\sum}}\frac{{logp}}{{p}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{6}}\underset{{p}} {\overset{\infty} {\sum}}\frac{{logp}}{{p}^{\mathrm{2}} −\mathrm{1}}=\frac{{log}\mathrm{1}}{\pi^{\mathrm{2}} }+\frac{{log}\mathrm{2}}{\mathrm{4}\pi^{\mathrm{2}} }+… \\ $$