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Question Number 94335 by mathmax by abdo last updated on 18/May/20

$$calculate\sum _{n=0}^{\mathrm{\infty }}{n}^{{(-1)}^n}{x}^n$$

Answered by mathmax by abdo last updated on 19/May/20

$$\mathrm{s}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\left(2\mathrm{n}\right){\mathrm{x}}^{2\mathrm{n}}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(2\mathrm{n}+1)}^{-1}{\mathrm{x}}^{2\mathrm{n}+1}$$$$=2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{nx}}^{2\mathrm{n}}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}+1}}{2\mathrm{n}+1}=\mathrm{h}\left(\mathrm{x}\right)+\mathrm{k}\left(\mathrm{x}\right)$$$$\mathrm{we}\mathrm{have}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}=\frac{1}{1-\mathrm{x}}\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}=\frac{1}{{(1-\mathrm{x})}^2}\Rightarrow$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}=\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{2\mathrm{n}}=\frac{{\mathrm{x}}^2}{{(1-{\mathrm{x}}^2)}^2}=\mathrm{h}\left(\mathrm{x}\right)$$$$\mathrm{k}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}+1}}{2\mathrm{n}+1}\Rightarrow {\mathrm{k}}^{{}^{\prime }}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{2\mathrm{n}}=\frac{1}{1-{\mathrm{x}}^2}\Rightarrow$$$$\mathrm{k}\left(\mathrm{x}\right)={\int }_0^{\mathrm{x}}\frac{\mathrm{dt}}{1-{\mathrm{t}}^2}+\mathrm{k}(\mathrm{k}=0)=\frac{1}{2}{\int }_0^{\mathrm{x}}(\frac{1}{1-\mathrm{t}}+\frac{1}{1+\mathrm{t}})\mathrm{dt}$$$$=\frac{1}{2}{[\ln \mid \frac{1+\mathrm{t}}{1-\mathrm{t}}\mid ]}_0^{\mathrm{x}}=\frac{1}{2}\ln \mid \frac{1+\mathrm{x}}{1-\mathrm{x}}\mid \Rightarrow \mathrm{s}\left(\mathrm{x}\right)=\frac{{2\mathrm{x}}^2}{{(1-{\mathrm{x}}^2)}^2}+\frac{1}{2}\ln \mid \frac{1+\mathrm{x}}{1-\mathrm{x}}\mid$$$$\mathrm{with}\mid \mathrm{x}\mid <1$$

Answered by maths mind last updated on 18/May/20

$$=\sum _{n⩾1}2{nx}^{2n}-\sum _{n⩾0}(2n+1)x^{2n+1}$$$$s=\sum _{n⩾1}2{nx}^{2n}-x\sum _{n⩾0}2{nx}^{2n}-x\sum _{n⩾0}{x}^{2n}$$$$\sum _{n⩾1}{nx}^{n-1}=\frac{1}{{(1-x)}^2}$$$$\Rightarrow \mathrm{\Sigma }2{nx}^{2n}=\frac{2x^2}{{(1-x^2)}^2}$$$$\Rightarrow s=\frac{2x^2}{{(1-x^2)}^2}-\frac{2x^3}{{(1-x^2)}^2}-\frac{x}{1-x^2}$$$$s=\frac{-x^3+2x^2-x}{{(1-x^2)}^2}$$

Commented by mathmax by abdo last updated on 19/May/20

$$\mathrm{thanks}\mathrm{sir}.$$

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