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Question Number 159080 by qaz last updated on 12/Nov/21
Σ_(n=0) ^∞ (1/(n!(n^4 +n^2 +1)))=?
∑∞n=01n!(n4+n2+1)=?
Answered by mindispower last updated on 13/Nov/21
=Σ_(n≥0) (1/(n!(n^2 +1−n)(n^2 +1+n))) =Σ_(n≥0) (1/(n!))(((an+b)/(n^2 −n+1))+((cn+d)/(n^2 +n+1))) a+c=0,b+d=1,(a+b−c+d)=0,a−c=−1 a=−(1/2),c=(1/2)=b=d =(1/2).Σ_(n≥0) (1/(n!))(.((−n+1)/(n^2 −n+1))+((n+1)/(n^2 +n+1))) =(1/2).Σ_(n≥0) .(1/(n!))(−(n/(n^2 −n+1))+(1/(n^2 +n+1))) =−(1/2)Σ_(n≥1) (1/((n−1)!(n^2 −n+1)_ ))∣_(=U_n ) +(1/2)Σ_(n≥0) (1/(n!(n^2 +n+1))), +(1/2)Σ_(n≥0) (1/(n!))((1/(n^2 −n+1))+(n/(n^2 +n+1))) ={−Σ_(n≥1) (1/2)U_n +Σ_(n≥0) (1/2)U_(n+1) }_(=0) +Σ_(n≥0) (1/(n!(n^2 −n+1)))+(1/2)Σ_(n≥0) (n/(n!(n^2 +n+1))) (n/(n!(n^2 +n+1)))=((n(n+1))/((n+1)!(n^2 +n+1)))=(1/((n+1)!))−(1/((n+1)!(n^2 +n+1))) =(1/2){Σ_(n≥0) (1/(n!(n^2 −n+1)_(.=V_n ) ))+Σ_(n≥0) (1/((n+1)!))−(1/((n+1)!(n^2 +n+1)))} =(1/2)Σ_(n≥0) V_n −Σ_(n≥0) V_(n+1) +(1/2)Σ_(n≥0) (1/((n+1)!))=(V_0 /2)+(1/2)(e−1) V_0 =1 S=(e/2)
=∑n⩾01n!(n2+1−n)(n2+1+n)=∑n⩾01n!(an+bn2−n+1+cn+dn2+n+1)a+c=0,b+d=1,(a+b−c+d)=0,a−c=−1a=−12,c=12=b=d=12.∑n⩾01n!(.−n+1n2−n+1+n+1n2+n+1)=12.∑n⩾0.1n!(−nn2−n+1+1n2+n+1)=−12∑n⩾11(n−1)!(n2−n+1)∣=Un+12∑n⩾01n!(n2+n+1),+12∑n⩾01n!(1n2−n+1+nn2+n+1)={−∑n⩾112Un+∑n⩾012Un+1}=0+∑n⩾01n!(n2−n+1)+12∑n⩾0nn!(n2+n+1)nn!(n2+n+1)=n(n+1)(n+1)!(n2+n+1)=1(n+1)!−1(n+1)!(n2+n+1)=12{∑n⩾01n!(n2−n+1).=Vn+∑n⩾01(n+1)!−1(n+1)!(n2+n+1)}=12∑n⩾0Vn−∑n⩾0Vn+1+12∑n⩾01(n+1)!=V02+12(e−1)V0=1S=e2
Commented by Tawa11 last updated on 13/Nov/21
Great sir
Greatsir
Commented by mindispower last updated on 16/Nov/21
thank you sir have nice day
thankyousirhaveniceday

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