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Question Number 33886 by math khazana by abdo last updated on 26/Apr/18

$$findthevalueof\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2n+3}.$$

Commented by math khazana by abdo last updated on 29/Apr/18

$$changementofindicen=p-1give$$$$S=\sum _{p=1}^{\mathrm{\infty }}\frac{{(-1)}^{p-1}}{2(p-1)+3}=\sum _{p=1}^{\mathrm{\infty }}\frac{{(-1)}^{p-1}}{2p+1}$$$$=-\sum _{p=1}^{\mathrm{\infty }}\frac{{(-1)}^p}{2p+1}=-\sum _{p=0}^{\mathrm{\infty }}\frac{{(-1)}^p}{2p+1}+1$$$$w\left(x\right)=\sum _{p=o}^{\mathrm{\infty }}\frac{{(-1)}^p}{2p+1}{x}^{2p+1}with\mid x\mid <1$$$$w^{{}^{\prime }}\left(x\right)=\sum _{p=0}^{\mathrm{\infty }}{(-1)}^p{x}^{2p}=\sum _{p=0}^{\mathrm{\infty }}{(-x^2)}^p=\frac{1}{1+x^2}$$$$\Rightarrow w\left(x\right)={\int }_0^x\frac{dt}{1+t^2}+\lambda but\lambda =w\left(0\right)=0\Rightarrow$$$$w\left(x\right)=arctanx$$$$S=1-w\left(1\right)=1-\frac{\pi }{4}$$$$★S=1-\frac{\pi }{4}★$$

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