Question Number 82447 by mathmax by abdo last updated on 21/Feb/20
$${calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\xi\left({n}\right) \\ $$
Answered by mind is power last updated on 21/Feb/20
$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}.\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{m}^{{n}} } \\ $$$$=\underset{{m}\geqslant\mathrm{1}} {\sum}\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{nm}^{{n}} } \\ $$$${ln}\left(\mathrm{1}+{x}\right)=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {x}^{{k}} }{{k}} \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}.\left(\frac{\mathrm{1}}{{m}}\right)^{{n}} =−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}.\left(\frac{\mathrm{1}}{{m}}\right)^{{n}} =−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{m}}\right) \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\left(\frac{\mathrm{1}}{{m}}\right)^{{n}} =−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{m}}\right)+\frac{\mathrm{1}}{{m}} \\ $$$$\underset{{m}\geqslant\mathrm{1}} {\sum}\left\{−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{m}}\right)+\frac{\mathrm{1}}{\mathrm{m}}\right\}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left\{−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{m}}\right)+\frac{\mathrm{1}}{{m}}\right\} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\:\underset{{m}=\mathrm{1}} {\overset{{x}} {\sum}}\left\{\frac{\mathrm{1}}{{m}}\right\}−{ln}\left(\mathrm{1}+{x}\right)=\gamma \\ $$
Commented by mathmax by abdo last updated on 21/Feb/20
$${thank}\:{you}\:{sir}\:. \\ $$
Commented by mind is power last updated on 21/Feb/20
$${withe}\:{pleasur} \\ $$