Question Number 155136 by amin96 last updated on 25/Sep/21
$$\boldsymbol{{prove}}\:\boldsymbol{{that}} \\ $$$$\frac{\boldsymbol{\mathrm{log}}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}}\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\varphi}\left(\boldsymbol{{n}}\right)}{\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\mathrm{2}} \mathrm{2}^{\boldsymbol{{n}}} }+\boldsymbol{\mathrm{log}}\left(\mathrm{2}\right)\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\varphi}\left({n}\right)}{\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\mathrm{3}} \mathrm{2}^{\boldsymbol{{n}}} }+\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\varphi}\left({n}\right)}{\left(\boldsymbol{{n}}+\mathrm{1}\right)^{\mathrm{4}} \mathrm{2}^{\boldsymbol{{n}}} }= \\ $$$$=\frac{\mathrm{23}}{\mathrm{8}}\boldsymbol{\zeta}\left(\mathrm{6}\right)−\mathrm{2}\boldsymbol{\zeta}^{\mathrm{2}} \left(\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{18}}\boldsymbol{\mathrm{log}}^{\mathrm{6}} \left(\mathrm{2}\right) \\ $$$$\boldsymbol{{m}}.\boldsymbol{{A}} \\ $$