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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)))=?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:...\:\:{calculus}\:.... \\ $$$$\:\:\:\:\:{find}\:::\:\:{i}::\:\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} −\mathrm{1}}=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ii}::\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{4}} −\mathrm{1}}\right)=? \\ $$$$\:\:\:\: \\ $$$$ \\ $$

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)

$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{15}}−\frac{\mathrm{1}}{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{35}}−\frac{\mathrm{1}}{\mathrm{48}}+... \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{35}}+\frac{\mathrm{1}}{\mathrm{63}}+..\right)−\left(\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{48}}+...\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}+...\right)−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{10}}+...\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{8}}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by mnjuly1970 last updated on 03/Feb/21

thanks  alot mr payan

$${thanks}\:\:{alot}\:{mr}\:{payan} \\ $$

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))

$$\mathrm{S}_{\mathrm{1}} \:=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{S}_{\mathrm{1}} \:=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}−\mathrm{1}\right)\left({n}+\mathrm{1}\right)} \\ $$$$\mathrm{S}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−\mathrm{1}}−\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\right] \\ $$$$\mathrm{S}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}−\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\right] \\ $$$$\mathrm{S}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}+\underset{{n}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\right] \\ $$$$\mathrm{S}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right]\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{S}_{\mathrm{2}} \:=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{4}} −\mathrm{1}} \\ $$$$\mathrm{S}_{\mathrm{2}} \:=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}^{\mathrm{2}} −\mathrm{1}\right)\left({n}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\mathrm{S}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} −\mathrm{1}}−\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$\mathrm{S}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{S}_{\mathrm{1}} −\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$\mathrm{S}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\left\{−\frac{{i}}{\mathrm{4}}\left[−\Psi\left(\mathrm{1}−\frac{{i}}{\mathrm{2}}\right)+\Psi\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)\right.\right. \\ $$$$\left.+\left.\Psi\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{{i}}{\mathrm{2}}\right)−\Psi\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{{i}}{\mathrm{2}}\right)\right]\right\} \\ $$$$\mathrm{S}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\left\{\frac{\pi}{\mathrm{2sinh}\pi}\right\} \\ $$$$\mathrm{S}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{8}}−\frac{\pi}{\mathrm{4sinh}\pi} \\ $$$$ \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 03/Feb/21

mercey and grateful mr olaf...

$${mercey}\:{and}\:{grateful}\:{mr}\:{olaf}... \\ $$

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