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Question Number 41349 by math khazana by abdo last updated on 06/Aug/18

$$calculate\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{3n-1}$$

Commented by maxmathsup by imad last updated on 06/Aug/18

$$letS=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{3n-1}changementofindicen=j+1give$$$$S=\sum _{j=0}^{\mathrm{\infty }}\frac{{(-1)}^{j+1}}{3j+2}=-\sum _{j=0}^{\mathrm{\infty }}\frac{{(-1)}^j}{3j+2}\Rightarrow S=-\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{3n+2}let$$$$w\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{3n+2}{x}^{3n+2}with\mid x\mid <1wehaveS=w\left(1\right)and$$$$w^{{}^{\prime }}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{x}^{3n+1}=x\sum _{n=0}^{\mathrm{\infty }}{(-x^3)}^n=\frac{x}{1+x^3}\Rightarrow$$$$w\left(x\right)=\int \frac{xdx}{1+x^3}+cletdecompseF\left(x\right)=\frac{x}{1+x^3}=\frac{x}{(x+1)(x^2-x+1)}$$$$F\left(x\right)=\frac{a}{x+1}+\frac{bx+c}{x^2-x+1}$$$$a={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{-1}{3}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+b\Rightarrow b=\frac{1}{3}\Rightarrow F\left(x\right)=-\frac{1}{3(x+1)}+\frac{1}{3}\frac{x+3c}{x^2-x+1}$$$$F\left(0\right)=0=-\frac{1}{3}+c\Rightarrow c=\frac{1}{3}\Rightarrow F\left(x\right)=\frac{-1}{3(x+1)}+\frac{1}{3}\frac{x+1}{x^2-x+1}\Rightarrow$$$$\int F\left(x\right)dx=-\frac{1}{3}ln\mid x+1\mid +\frac{1}{6}\int \frac{2x-1+3}{x^2-x+1}dx$$$$=-\frac{1}{3}ln\mid x+1\mid +\frac{1}{6}ln(x^2-x+1)+\frac{1}{2}\int \frac{dx}{{(x-\frac{1}{2})}^2+\frac{3}{4}}but$$$$\int \frac{dx}{{(x-\frac{1}{2})}^2+\frac{3}{4}}{=}_{x-\frac{1}{2}=\frac{\sqrt{3}}{2}t}\frac{4}{3}\int \frac{1}{1+t^2}\frac{\sqrt{3}}{2}dt$$$$=\frac{2}{\sqrt{3}}arctan\left(\frac{2x-1}{\sqrt{3}}\right)\Rightarrow w\left(x\right)=-\frac{1}{3}ln\mid x+1\mid +\frac{1}{6}ln(x^2-x+1)+\frac{2}{\sqrt{3}}arctan\left(\frac{2x-1}{\sqrt{3}}\right)+c$$$$w\left(0\right)=0=-\frac{2}{\sqrt{3}}arctan\left(\frac{1}{\sqrt{3}}\right)+c\Rightarrow c=\frac{2}{\sqrt{3}}arctan\left(\frac{1}{\sqrt{3}}\right)=\frac{2}{\sqrt{3}}\frac{\pi }{6}=\frac{\pi }{3\sqrt{3}}\Rightarrow$$$$w\left(x\right)=-\frac{1}{3}ln\mid x+1\mid +\frac{1}{6}ln(x^2-x+1)+\frac{2}{\sqrt{3}}arctan\left(\frac{2x-1)}{\sqrt{3}}\right)+\frac{\pi }{3\sqrt{3}}$$$$S=w\left(1\right)=-\frac{1}{3}ln\left(2\right)+\frac{2}{\sqrt{3}}arctan\left(\frac{1}{\sqrt{3}}\right)+\frac{\pi }{3\sqrt{3}}=-\frac{ln\left(2\right)}{3}+\frac{2}{\sqrt{3}}\frac{\pi }{6}+\frac{\pi }{3\sqrt{3}}$$$$S=\frac{2\pi }{3\sqrt{3}}-\frac{ln\left(2\right)}{3}.$$$$$$

Commented by math khazana by abdo last updated on 07/Aug/18

$$S=-w\left(1\right)\Rightarrow S=\frac{ln\left(2\right)}{3}-\frac{2\pi }{3\sqrt{3}}.$$

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