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Question Number 92925 by frc2crc last updated on 09/May/20

$$S_n=\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(4k^2-1)}^n}$$$$findasimplerform$$

Commented by mr W last updated on 09/May/20

$$checkyourquestionplease.$$$$\underset{k=1}{\sum ^{nor\mathrm{\infty }}}$$

Commented by frc2crc last updated on 09/May/20

$$\underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(4k^2-1)}^n}thissumforanyn>0$$

Commented by prakash jain last updated on 09/May/20

$$\mathrm{series}\mathrm{is}\mathrm{convergent}\mathrm{for}n⩾1.\mathrm{We}$$$$\mathrm{can}\mathrm{use}\mathrm{limit}\mathrm{comparion}\mathrm{test}.$$$$n=1$$$$\frac{1}{4k^2-1}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1})$$$$\Rightarrow \underset{k=1}{\sum ^{\mathrm{\infty }}}\frac{1}{4k^2-1}=\frac{1}{2}$$$$\mathrm{n}=2\mathrm{can}\mathrm{be}\mathrm{evaluated}\mathrm{using}$$$$\mathrm{known}\mathrm{series}\mathrm{expansion}\mathrm{for}{\pi }^2.$$$$-----$$$$\mathrm{I}\mathrm{will}\mathrm{check}\mathrm{if}\mathrm{a}\mathrm{generic}\mathrm{closed}$$$$\mathrm{form}\mathrm{formula}\mathrm{is}\mathrm{possible}.$$$$\mathrm{It}\mathrm{may}\mathrm{exists}\mathrm{using}\mathrm{zeta}\mathrm{function}.$$$$$$

Commented by abdomathmax last updated on 10/May/20

$$lettakeatryletdecomposeF\left(x\right)=\frac{1}{{(4x^2-1)}^n}$$$$=\frac{1}{{(2x-1)}^n{(2x+1)}^n}changement2x-1=tgive$$$$x=\frac{t+1}{2}\Rightarrow 2x+1=t+2\Rightarrow F\left(x\right)=g\left(t\right)=\frac{1}{t^n{(t+2)}^n}$$$$g\left(t\right)=\sum _{i=1}^n\frac{a_i}{t^i}+\sum _{i=1}^n\frac{b_i}{{(t+2)}^i}whatis{a}_i?$$$$wefind{D}_{n-1}\left(0\right)forh\left(t\right)={(t+2)}^{-n}bytaylor$$$$seriwegeth\left(t\right)=\sum _{p=0}^{n-1}\frac{h^{\left(p\right)}\left(0\right)}{\left(p\right)!}{t}^p+\frac{t^n}{n!}\xi \left(t\right)$$$$h^{,}\left(t\right)=-n{(t+2)}^{-n-1}$$$$h^{\left(2\right)}\left(t\right)=-n(-n-1){(t+2)}^{-n-2}={(-1)}^{2}n(n+1){(t+2)}^{-n-2}$$$$h^{\left(p\right)}\left(x\right)={(-1)}^{p}n(n+1)...(n+p-1){(t+2)}^{-n-p}$$$$\Rightarrow h^{\left(p\right)}\left(0\right)={(-1)}^{p}n(n+1)....(n+p-1)2^{-n-p}\Rightarrow$$$$h\left(t\right)=\sum _{p=0}^{n-1}\frac{{(-1)}^{p}n(n+1)...(n+p-1)}{p!2^{n+p}}{t}^p+\frac{t^n}{n!}\xi \left(t\right)\Rightarrow$$$$g\left(t\right)=\sum _{p=0}^{n-1}\frac{{(-1)}^{p}n(n+1)...(n+p-1)}{p!2^{n+p}{t}^{n-p}}+\frac{1}{n!}\xi \left(t\right)$$$${=}_{n-p=i}\sum _{i=1}^n\frac{{(-1)}^{n-i}n(n+1)...(2n-i-1)}{(n-i)!2^{2n-i}{t}^i}+...$$$$\Rightarrow a_i=\frac{{(-1)}^{n-i}n(n+1)....(2n-i-1)}{(n-i)!\times 2^{2n-i}}$$$$whatis{b}_i?wedetermine{D}_{n-1}(-2)forh\left(t\right)=t^{-n}$$$$h\left(t\right)=\sum _{p=0}^{n-1}\frac{h^{\left(p\right)}(-2)}{p!}{(t+2)}^p+\frac{{(t+2)}^n}{n!}\xi \left(t\right)$$$$h^{\left(p\right)}\left(t\right)={(-1)}^{p}n(n+1)...(n+p-1)t^{-n-p}$$$$\Rightarrow h^{\left(p\right)}(-2)={(-1)}^{p}n(n+1)...(n+p-1){(-2)}^{-n-p}\Rightarrow$$$$h\left(t\right)=\sum _{p=0}^{n-1}\frac{{(-1)}^{p}n(n+1)...(n+p-1){(-2)}^{-n-p}}{p!}{(t+2)}^p+...$$$$\Rightarrow g\left(t\right)=\sum _{p=0}^{n-1}\frac{{(-1)}^{p}n(n+1)...(n+p-1)}{p!{(-2)}^{n+p}{(t+2)}^{n-p}}+\frac{1}{n!}\xi \left(t\right)$$$${=}_{n-p=i}\sum _{i=1}^n\frac{{(-1)}^{n-i}n(n+1)...(2n-i-1)}{(n-i)!{(-2)}^{2n-i}{(t+2)}^i}+...\Rightarrow$$$$b_i=\frac{{(-1)}^{n-i}n(n+1)...(2n-i-1)}{(n-i)!{(-2)}^{2\mathit{n}-\mathit{i}}}$$$$...\mathit{b}\mathit{e}\mathit{c}\mathit{o}\mathit{n}\mathit{t}\mathit{i}\mathit{n}\mathit{u}\mathit{e}\mathit{d}....$$

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