Question Number 183827 by SANOGO last updated on 30/Dec/22
$${calcul}:\:\:{S}=\underset{{n}=\mathrm{2}\:\:} {\overset{+{oo}} {\sum}}\frac{{n}}{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Answered by Ar Brandon last updated on 30/Dec/22
$${S}=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{n}}{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{n}}{\left(\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{{n}\left(\left({n}+\mathrm{1}\right)−\left({n}−\mathrm{1}\right)\right)^{\mathrm{2}} }{\left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{{n}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{2}{n}}{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)}+\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}+\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{2}} {\sum}\left(\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)=\frac{\mathrm{5}}{\mathrm{16}} \\ $$
Commented by MJS_new last updated on 30/Dec/22
$$\mathrm{nice}! \\ $$
Commented by Ar Brandon last updated on 30/Dec/22
Thanks master��
Commented by SANOGO last updated on 30/Dec/22
$${merci} \\ $$
Commented by MJS_new last updated on 31/Dec/22
I'm still a beginner, maybe with extra years of experience ��