Question Number 114099 by mnjuly1970 last updated on 17/Sep/20

   ....  mathematical  analysis....      prove that ::                                                   Σ_(n=1) ^∞ ( ((3^n −1)/4^n ))ζ(n+1) =π                 m.n.july.1970#

Answered by maths mind last updated on 17/Sep/20

=Σ_(n≥1) .Σ_(m≥1) (((3^n −1)/4^n )).(1/m^(n+1) )  =Σ_(m≥1) (1/m){Σ_(n≥1) ((3/(4.m)))^n −Σ_(n≥1) ((1/(4m)))^n }  =Σ_(m≥1) (1/m){(3/(4m)).(1/(1−(3/(4m))))−(1/(4m)).(1/(1−(1/(4m))))}  =Σ_(m≥1) (1/m){(3/(4m−3))}−Σ_(m≥1) (1/(m(4m−1)))  =Σ_(m≥0) (3/((1+m)(4m+1)))−Σ_(m≥0) (1/((m+1)(4m+3)))  =Σ_(m≥0) ((1−(1/4))/((m+1)(m+(1/4))))−Σ_(m≥0) ((1−(3/4))/((m+1)(m+(3/4))))  =Ψ(1)−Ψ((1/4))−{Ψ(1)−Ψ((3/4))}  =Ψ((3/4))−Ψ((1/4))=Ψ(1−(1/4))−Ψ((1/4))=πcot((π/4))=π

Commented bymnjuly1970 last updated on 18/Sep/20

thank you sir.grateful...