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Question Number 82445 by mathmax by abdo last updated on 21/Feb/20
$${calculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{\xi\left({n}\right)−\mathrm{1}}{{n}} \\ $$ $${with}\:\xi\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{{x}} }\:\:\:\left({x}>\mathrm{1}\right) \\ $$
Answered by mind is power last updated on 21/Feb/20
![=Σ_(n≥2) (1/n).Σ_(m≥2) (1/m^n )=Σ_(m≥2) .Σ_(n≥2) (1/n).((1/m))^n =Σ_(m≥2) .[−ln(1−(1/m))−(1/m)] =Σ_(m≥2) [−ln(m−1)+ln(m)−(1/m)] =lim_(x→∞) .Σ_(m=2) ^x −ln(m−1)+ln(m)−(1/m)] =lim_(x→∞) ln(x)−Σ_(m=2) ^x (1/m) =−lim_(x→∞) {Σ_(m=2) ^x (1/m)−ln(x)} =−lim_(x→∞) {Σ_(m=1) ^x (1/m)−ln(x)−1}=−{γ−1}=1−γ γ Euler mechorini constent](Q82478.png)
$$=\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{{n}}.\underset{{m}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{{m}^{{n}} }=\underset{{m}\geqslant\mathrm{2}} {\sum}.\underset{{n}\geqslant\mathrm{2}} {\sum}\frac{\mathrm{1}}{{n}}.\left(\frac{\mathrm{1}}{{m}}\right)^{{n}} =\underset{{m}\geqslant\mathrm{2}} {\sum}.\left[−{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{m}}\right)−\frac{\mathrm{1}}{{m}}\right] \\ $$ $$=\underset{{m}\geqslant\mathrm{2}} {\sum}\left[−{ln}\left({m}−\mathrm{1}\right)+{ln}\left({m}\right)−\frac{\mathrm{1}}{{m}}\right] \\ $$ $$\left.=\underset{{x}\rightarrow\infty} {\mathrm{lim}}.\underset{{m}=\mathrm{2}} {\overset{{x}} {\sum}}−{ln}\left({m}−\mathrm{1}\right)+{ln}\left({m}\right)−\frac{\mathrm{1}}{{m}}\right] \\ $$ $$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{ln}\left({x}\right)−\underset{{m}=\mathrm{2}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{{m}} \\ $$ $$=−\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left\{\underset{{m}=\mathrm{2}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{{m}}−{ln}\left({x}\right)\right\} \\ $$ $$=−\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left\{\underset{{m}=\mathrm{1}} {\overset{{x}} {\sum}}\frac{\mathrm{1}}{{m}}−{ln}\left({x}\right)−\mathrm{1}\right\}=−\left\{\gamma−\mathrm{1}\right\}=\mathrm{1}−\gamma\: \\ $$ $$\gamma\:{Euler}\:{mechorini}\:{constent} \\ $$