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Question Number 82447 by mathmax by abdo last updated on 21/Feb/20

calculate Σ_(n=2) ^∞ (((−1)^n )/n)ξ(n)

$${calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\xi\left({n}\right) \\ $$

Answered by mind is power last updated on 21/Feb/20

=Σ_(n≥2) (((−1)^n )/n).Σ_(m≥1) (1/m^n ) =Σ_(m≥1) Σ_(n≥2) (((−1)^n )/(nm^n )) ln(1+x)=Σ_(k≥1) (((−1)^(k−1) x^k )/k) ⇒Σ_(n≥1) (((−1)^n )/n).((1/m))^n =−Σ_(n≥1) (((−1)^(n−1) )/n).((1/m))^n =−ln(1+(1/m)) ⇒Σ_(n≥2) (((−1)^n )/n)((1/m))^n =−ln(1+(1/m))+(1/m) Σ_(m≥1) {−ln(1+(1/m))+(1/m)}=lim_(x→∞) {−ln(1+(1/m))+(1/m)} =lim_(x→∞) Σ_(m=1) ^x {(1/m)}−ln(1+x)=γ

$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}.\underset{{m}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{m}^{{n}} } \\ $$$$=\underset{{m}\geqslant\mathrm{1}} {\sum}\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{nm}^{{n}} } \\ $$$${ln}\left(\mathrm{1}+{x}\right)=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {x}^{{k}} }{{k}} \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}.\left(\frac{\mathrm{1}}{{m}}\right)^{{n}} =−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}.\left(\frac{\mathrm{1}}{{m}}\right)^{{n}} =−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{m}}\right) \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\left(\frac{\mathrm{1}}{{m}}\right)^{{n}} =−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{m}}\right)+\frac{\mathrm{1}}{{m}} \\ $$$$\underset{{m}\geqslant\mathrm{1}} {\sum}\left\{−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{m}}\right)+\frac{\mathrm{1}}{\mathrm{m}}\right\}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left\{−{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{m}}\right)+\frac{\mathrm{1}}{{m}}\right\} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\:\underset{{m}=\mathrm{1}} {\overset{{x}} {\sum}}\left\{\frac{\mathrm{1}}{{m}}\right\}−{ln}\left(\mathrm{1}+{x}\right)=\gamma \\ $$

Commented by mathmax by abdo last updated on 21/Feb/20

thank you sir .

$${thank}\:{you}\:{sir}\:. \\ $$

Commented by mind is power last updated on 21/Feb/20

withe pleasur

$${withe}\:{pleasur} \\ $$

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