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evaluate-n-0-1-n-n-4-n-2-1-e-2-

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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)
evaluate::∑∞n=01n!(n4+n2+1)=e2
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)
n4+n2+1=n4+2n2+1−n2=(n2+1)2−n2=(n2+n+1)(n2−n+1)1(n2+n+1)(n2−n+1)=12n(.1n2−n+1−1n2+n+1)S=∑n⩾01n!(n4+n2+1)∑n⩾0f(n)=1+∑n⩾1f(n)=1+∑n⩾112n(n)!.(1n2−n+1−1n2+n+1)=S∑n⩾11(2n)(n)!.1n2−n+1=12+∑n⩾112(n+1).(n+1)!.1(n2+n+1)S=32+∑n⩾112(n+1)2n!(n2+n+1)−12n(n2+n+1)n!=32+∑n⩾11(n2+n+1)n!(12(n+1)2−12n)32+Σ(n−(n+1)22n(n+1)2).1n2+n+1.1n!=32−12.∑n⩾1n2+n+1n(n+1).1n!(n2+n+1)=32−12∑n⩾11n(n+1)(n+1)!∑n⩾11n(n+1)(n+1)!=∑n⩾11n(n+1)!−1(n+1)(n+1)!Missing \left or extra \right−∑n⩾11(n+1)!=1−(e−2)=3−eS=32−12(3−e)=e2

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