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Question Number 81206 by Power last updated on 10/Feb/20

Commented by mathmax by abdo last updated on 10/Feb/20

$$letf\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)}{x}^{n}with\mid x\mid ⩽1andx\ne -1$$$$f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)}{x}^n-1=\frac{1}{\sqrt{x}}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)}{\left(\sqrt{x}\right)}^{2n+1}-1$$$$=\frac{1}{\sqrt{x}}\phi \left(\sqrt{x}\right)-1with\phi \left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)}{x}^{2n+1}$$$$wehave{\phi }^{{}^{\prime }}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^{2n}=\sum _{n=0}^{\mathrm{\infty }}{(-x^2)}^n=\frac{1}{1+x^2}\Rightarrow$$$$\phi \left(x\right)=\int \frac{dx}{1+x^2}+c=arctan\left(x\right)+c(c=\phi \left(0\right)=0)\Rightarrow \phi \left(x\right)=arctan\left(x\right)$$$$\Rightarrow f\left(x\right)=\frac{1}{\sqrt{x}}arctan\left(\sqrt{x}\right)-1and$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)3^n}=f\left(\frac{1}{3}\right)=\sqrt{3}arctan\left(\frac{1}{\sqrt{3}}\right)-1$$$$=\sqrt{3}\times \frac{\pi }{6}-1=\frac{\pi \sqrt{3}}{6}-1.$$$$$$

Answered by mind is power last updated on 10/Feb/20

$$arctan\left(x\right)=\sum _{n⩾0}\frac{{(-1)}^n{x}^{2n+1}}{(2n+1)}$$$$\Rightarrow \frac{arctan\left(x\right)}{x}=1+\sum _{n⩾1}\frac{{(-1)}^n{x}^{2n}}{2n+1}$$$$x=\frac{1}{\sqrt{3}}\Rightarrow \sqrt{3}arctan\left(\frac{1}{\sqrt{3}}\right)-1=\sum _{n⩾1}\frac{{(-1)}^n}{(2n+1)3^n}$$$$\iff \sum _{n⩾1}\frac{{(-1)}^n}{(2n+1)3^n}=\sqrt{3}.\frac{\pi }{6}-1=\frac{\pi }{2\sqrt{3}}-1$$$$$$$$$$

Commented by Power last updated on 10/Feb/20

$$\mathrm{thanks}$$

Commented by mind is power last updated on 10/Feb/20

$$withepleasur$$

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