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calculus-find-i-n-2-1-n-n-2-1-ii-n-2-1-n-n-4-1-




Question Number 131286 by mnjuly1970 last updated on 03/Feb/21
             ...  calculus ....       find ::  i::  Σ_(n=2) ^∞ (((−1)^n )/(n^2 −1))=?                       ii:: Σ_(n=2) ^∞ ((((−1)^n )/(n^4 −1)))=?
calculus.find::i::n=2(1)nn21=?ii::n=2((1)nn41)=?
Answered by Dwaipayan Shikari last updated on 03/Feb/21
Σ_(n=2) ^∞ (((−1)^n )/(n^2 −1))=(1/3)−(1/8)+(1/(15))−(1/(24))+(1/(35))−(1/(48))+...  =((1/3)+(1/(15))+(1/(35))+(1/(63))+..)−((1/8)+(1/(24))+(1/(48))+...)  =(1/2)(1−(1/3)+(1/3)−(1/5)+...)−(1/8)(1+(1/3)+(1/6)+(1/(10))+...)  =(1/2)−(1/8)(Σ_(n=1) ^∞ (2/n)−(1/(n+1)))  =(1/2)−(1/4)=(1/4)
n=2(1)nn21=1318+115124+135148+=(13+115+135+163+..)(18+124+148+)=12(113+1315+)18(1+13+16+110+)=1218(n=12n1n+1)=1214=14
Commented by mnjuly1970 last updated on 03/Feb/21
thanks  alot mr payan
thanksalotmrpayan
Answered by Olaf last updated on 03/Feb/21
S_1  = Σ_(n=2) ^∞ (((−1)^n )/(n^2 −1))  S_1  = Σ_(n=2) ^∞ (((−1)^n )/((n−1)(n+1)))  S_1  = (1/2)[Σ_(n=2) ^∞ (((−1)^n )/(n−1))−Σ_(n=2) ^∞ (((−1)^n )/(n+1))]  S_1  = (1/2)[Σ_(n=1) ^∞ (((−1)^(n+1) )/n)−Σ_(n=3) ^∞ (((−1)^(n−1) )/n)]  S_1  = (1/2)[−Σ_(n=1) ^∞ (((−1)^n )/n)+Σ_(n=3) ^∞ (((−1)^n )/n)]  S_1  = (1/2)[1−(1/2)] = (1/4)    S_2  = Σ_(n=2) ^∞ (((−1)^n )/(n^4 −1))  S_2  = Σ_(n=2) ^∞ (((−1)^n )/((n^2 −1)(n^2 +1)))  S_2  = (1/2)[Σ_(n=2) ^∞ (((−1)^n )/(n^2 −1))−Σ_(n=2) ^∞ (((−1)^n )/(n^2 +1))]  S_2  = (1/2)[S_1 −Σ_(n=2) ^∞ (((−1)^n )/(n^2 +1))]  S_2  = (1/8)−(1/2){−(i/4)[−Ψ(1−(i/2))+Ψ(1+(i/2))  +Ψ((3/2)−(i/2))−Ψ((3/2)+(i/2))]}  S_2  = (1/8)−(1/2){(π/(2sinhπ))}  S_2  = (1/8)−(π/(4sinhπ))
S1=n=2(1)nn21S1=n=2(1)n(n1)(n+1)S1=12[n=2(1)nn1n=2(1)nn+1]S1=12[n=1(1)n+1nn=3(1)n1n]S1=12[n=1(1)nn+n=3(1)nn]S1=12[112]=14S2=n=2(1)nn41S2=n=2(1)n(n21)(n2+1)S2=12[n=2(1)nn21n=2(1)nn2+1]S2=12[S1n=2(1)nn2+1]S2=1812{i4[Ψ(1i2)+Ψ(1+i2)+Ψ(32i2)Ψ(32+i2)]}S2=1812{π2sinhπ}S2=18π4sinhπ
Commented by mnjuly1970 last updated on 03/Feb/21
mercey and grateful mr olaf...
merceyandgratefulmrolaf