Question Number 148020 by qaz last updated on 25/Jul/21
$$\underset{\mathrm{k}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\mathrm{k}} \centerdot\frac{\mathrm{lnk}}{\mathrm{k}}=\gamma\mathrm{ln2}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \mathrm{2} \\ $$