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Find-out-A-n-2-n-n-3-n-where-p-n-1-1-n-p-

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Question Number 78490 by ~blr237~ last updated on 18/Jan/20
Find out A =Σ_(n=2) ^∞ ((ζ(n))/(n(−3)^n )) where ζ(p)=Σ_(n=1) ^∞ (1/n^p )
FindoutA=∑∞n=2ζ(n)n(−3)nwhereζ(p)=∑∞n=11np
Answered by mind is power last updated on 19/Jan/20
let f(x)=Σ_(n≥2) ((ζ(n)x^n )/n)=Σ_(n≥2) (x^n /n)Σ_(m≥1) (1/m^n ) =Σ_(n≥2) .Σ_(m≥1) ((x/m))^n .(1/n) ∀n,m ,n≥2 and m≥1 we haveΣ_(n≥2) Σ_(m≥1) ((x/m))^n .(1/n),≤Σ_(n≥2) (x^n /n).ζ(2)<∞ ,∀x∈]−1,1[ ⇒Σ_(n≥2) .Σ_(m≥1) ((x/m))^n .(1/n)=Σ_(m≥1) Σ_(n≥2) ((((x/m))^n )/n)=f(x) Σ_(k≥1) (a^k /k)=−ln(1−a)⇒Σ_(k≥2) (a^k /k)=−ln(1−a)−a f(x)=Σ_(m≥1) {−ln(1−(x/m))−(x/m)}=f(x) we have our sum=f(−(1/3)) Γ(x)=(1/x)Π_(k≥1) (((1+(1/k))^x )/(1+(x/k)))⇒(1/x)Π_(k≥1) .(e^(xln(1+(1/k))) /(1+(x/k)))=(1/x).Π_(k≥1) (e^(xln(1+(1/k))−(x/k)+(x/k)) /(1+(x/k))) =(1/x).Π_(k≥1) (e^(x(ln(1+(1/k))−(1/k))+(x/k)) /(1+(x/k)))=(1/x).e^(Σ_(k≥1) x(ln(1+(1/k))−(1/k))) .Π_(k≥1) (e^(x/k) /e^(ln(1+(x/k))) ) Σ_(k≥1) {ln(1+(1/k))−(1/k)}=−γ Euler macheronie Constent Γ(x)=(e^(−γx) /x).e^(Σ_(k≥1) {(x/k)−ln(1+(x/k))}) ⇒lnΓ(x)=−γx−ln(x)+Σ_(k≥1) {(x/k)−ln(1+(x/k))} ⇒Σ_(k≥1) {(x/k)−ln(1+(x/k))}=ln(Γ(x))+γx+ln(x) ⇒Σ_(k≥1) {−(x/k)−ln(1−(x/k))}=f(x)=ln(Γ(−x))−γx+ln(−x) S=f(−(1/3))=ln(Γ((1/3)))+(γ/3)+ln((1/3))=Σ_(n≥2) ((ζ(n))/(n(−3)^n ))
letf(x)=∑n⩾2ζ(n)xnn=∑n⩾2xnn∑m⩾11mn=∑n⩾2.∑m⩾1(xm)n.1n∀n,m,n⩾2andm⩾1wehave∑n⩾2∑m⩾1(xm)n.1n,⩽∑n⩾2xnn.ζ(2)<∞,∀x∈]−1,1[⇒∑n⩾2.∑m⩾1(xm)n.1n=∑m⩾1∑n⩾2(xm)nn=f(x)∑k⩾1akk=−ln(1−a)⇒∑k⩾2akk=−ln(1−a)−af(x)=∑m⩾1{−ln(1−xm)−xm}=f(x)wehaveoursum=f(−13)Γ(x)=1x∏k⩾1(1+1k)x1+xk⇒1x∏k⩾1.exln(1+1k)1+xk=1x.∏k⩾1exln(1+1k)−xk+xk1+xk=1x.∏k⩾1ex(ln(1+1k)−1k)+xk1+xk=1x.e∑k⩾1x(ln(1+1k)−1k).∏k⩾1exkeln(1+xk)∑k⩾1{ln(1+1k)−1k}=−γEulermacheronieConstentΓ(x)=e−γxx.e∑k⩾1{xk−ln(1+xk)}⇒lnΓ(x)=−γx−ln(x)+∑k⩾1{xk−ln(1+xk)}⇒∑k⩾1{xk−ln(1+xk)}=ln(Γ(x))+γx+ln(x)⇒∑k⩾1{−xk−ln(1−xk)}=f(x)=ln(Γ(−x))−γx+ln(−x)S=f(−13)=ln(Γ(13))+γ3+ln(13)=∑n⩾2ζ(n)n(−3)n

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