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Question Number 145200 by qaz last updated on 03/Jul/21
evaluate:: Σ_(n=0) ^∞ (1/(n!(n^4 +n^2 +1)))=(e/2)
$$\mathrm{evaluate}::\:\:\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{n}!\left(\mathrm{n}^{\mathrm{4}} +\mathrm{n}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\mathrm{e}}{\mathrm{2}} \\ $$
Answered by mindispower last updated on 03/Jul/21
n^4 +n^2 +1=n^4 +2n^2 +1−n^2 =(n^2 +1)^2 −n^2 =(n^2 +n+1)(n^2 −n+1) (1/((n^2 +n+1)(n^2 −n+1)))=(1/(2n))(.(1/(n^2 −n+1))−(1/(n^2 +n+1))) S=Σ_(n≥0) (1/(n!(n^4 +n^2 +1)))Σ_(n≥0) f(n)=1+Σ_(n≥1) f(n) =1+Σ_(n≥1) (1/(2n(n)!)).((1/(n^2 −n+1))−(1/(n^2 +n+1)))=S Σ_(n≥1) (1/((2n)(n)!)).(1/(n^2 −n+1)) =(1/2)+Σ_(n≥1) (1/(2(n+1).(n+1)!)).(1/((n^2 +n+1))) S=(3/2)+Σ_(n≥1) (1/(2(n+1)^2 n!(n^2 +n+1)))−(1/(2n(n^2 +n+1)n!)) =(3/2)+Σ_(n≥1) (1/((n^2 +n+1)n!))((1/(2(n+1)^2 ))−(1/(2n))) (3/2)+Σ(((n−(n+1)^2 )/(2n(n+1)^2 ))).(1/(n^2 +n+1)).(1/(n!)) =(3/2)−(1/2).Σ_(n≥1) ((n^2 +n+1)/(n(n+1))).(1/(n!(n^2 +n+1))) =(3/2)−(1/2)Σ_(n≥1) (1/(n(n+1)(n+1)!)) Σ_(n≥1) (1/(n(n+1)(n+1)!))=Σ_(n≥1) (1/(n(n+1)!))−(1/((n+1)(n+1)!)) =Σ_(n≥1) ((n+1−n)/(n(n+1)!))−(1/((n+1)(n+1)!))=Σ_(n≥1) ((1/(n(n)!))−(1/((n+1)(n+1)!)))_(telescopic) −Σ_(n≥1) (1/((n+1)!)) =1−(e−2)=3−e S=(3/2)−(1/2)(3−e)=(e/2)
$${n}^{\mathrm{4}} +{n}^{\mathrm{2}} +\mathrm{1}={n}^{\mathrm{4}} +\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}−{n}^{\mathrm{2}} =\left({n}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −{n}^{\mathrm{2}} \\ $$$$=\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)\left({n}^{\mathrm{2}} −{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}{n}}\left(.\frac{\mathrm{1}}{{n}^{\mathrm{2}} −{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right) \\ $$$${S}=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}!\left({n}^{\mathrm{4}} +{n}^{\mathrm{2}} +\mathrm{1}\right)}\underset{{n}\geqslant\mathrm{0}} {\sum}{f}\left({n}\right)=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}{f}\left({n}\right) \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{2}{n}\left({n}\right)!}.\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} −{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right)={S} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)\left({n}\right)!}.\frac{\mathrm{1}}{{n}^{\mathrm{2}} −{n}+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right).\left({n}+\mathrm{1}\right)!}.\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)} \\ $$$${S}=\frac{\mathrm{3}}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} {n}!\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}{n}\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right){n}!} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right){n}!}\left(\frac{\mathrm{1}}{\mathrm{2}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}{n}}\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}+\Sigma\left(\frac{{n}−\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right).\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}.\frac{\mathrm{1}}{{n}!} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}.\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{n}^{\mathrm{2}} +{n}+\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}.\frac{\mathrm{1}}{{n}!\left({n}^{\mathrm{2}} +{n}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)!} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)!}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)!} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{n}+\mathrm{1}−{n}}{{n}\left({n}+\mathrm{1}\right)!}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)!}=\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}\left({n}\right)!}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{1}\right)!}\underset{{telescopic}} {\right)} \\ $$$$−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$$$=\mathrm{1}−\left({e}−\mathrm{2}\right)=\mathrm{3}−{e} \\ $$$${S}=\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}−{e}\right)=\frac{{e}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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