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Question Number 72889 by mathmax by abdo last updated on 04/Nov/19

$$calculate\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)n^2}$$

Commented by mathmax by abdo last updated on 04/Nov/19

$$letS=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2(2n+1)}firstletdecomposeF\left(x\right)=\frac{1}{x^2(2x+1)}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{2x+1}$$$$b={lim}_{x\to 0}{x}^{2}F\left(x\right)=1$$$$c={lim}_{x\to -\frac{1}{2}}(2x+1)F\left(x\right)=4\Rightarrow F\left(x\right)=\frac{a}{x}+\frac{1}{x^2}+\frac{4}{2x+1}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+\frac{c}{2}\Rightarrow a=-2\Rightarrow F\left(x\right)=-\frac{2}{x}+\frac{1}{x^2}+\frac{4}{2x+1}\Rightarrow$$$$S=\sum _{n=1}^{\mathrm{\infty }}{(-1)}^{n}F\left(n\right)=-2\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}+\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}+4\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n}=-ln\left(2\right)$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}=(2^{1-2}-1)\xi \left(2\right)=-\frac{{\pi }^2}{12}$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+1}-1=\frac{\pi }{4}-1\Rightarrow$$$$S=-2(-ln2)-\frac{{\pi }^2}{12}+4(\frac{\pi }{4}-1)=2ln\left(2\right)-\frac{{\pi }^2}{12}+\pi -4$$

Answered by mind is power last updated on 04/Nov/19

$$\frac{1}{(2\mathrm{n}+1){\mathrm{n}}^2}=\frac{1}{{\mathrm{n}}^2}+\frac{\mathrm{c}}{\mathrm{n}}+\frac{4}{2\mathrm{n}+1}\Rightarrow \frac{1}{{\mathrm{n}}^2}+\frac{-2}{{\mathrm{n}}^{}}+\frac{4}{2\mathrm{n}+1}$$$$\sum _{\mathrm{n}⩾1}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}+\sum _{\mathrm{n}⩾1}\frac{{(-1)}^{\mathrm{n}+1}}{\mathrm{n}}+4\mathrm{\Sigma }\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}$$$$\frac{1}{1+{\mathrm{x}}^2}=\sum _{\mathrm{n}⩾0}{(-{\mathrm{x}}^2)}^{\mathrm{n}}$$$$\arctan \left(\mathrm{x}\right)=\mathrm{\Sigma }{(-1)}^{\mathrm{n}}.\frac{{\mathrm{x}}^{2\mathrm{n}+1}}{2\mathrm{n}+1}$$$$\mathrm{x}=1\Rightarrow \mathrm{\Sigma }\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}=\frac{\pi }{4}-1$$$$\sum _{\mathrm{n}⩾1}\frac{{(-1)}^{\mathrm{n}+1}}{\mathrm{n}}\to \ln \left(2\right)$$$$\sum _{\mathrm{n}⩾1}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}=\frac{1}{4}.\zeta \left(2\right)-\frac{3}{4}\zeta \left(2\right)=-\frac{\zeta \left(2\right)}{2}=-\frac{{\pi }^2}{12}$$$$\Rightarrow \sum _{\mathrm{n}⩾1}\frac{{(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1){\mathrm{n}}^2}=-\frac{{\pi }^2}{12}+2\ln \left(2\right)+\pi -4$$

Commented by turbo msup by abdo last updated on 04/Nov/19

$$thsnkyousir.$$

Commented by mind is power last updated on 04/Nov/19

$$\mathrm{y},\mathrm{re}\mathrm{welcom}$$

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