Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 31982 by abdo imad last updated on 17/Mar/18

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

Commented by 6123 last updated on 18/Mar/18

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{(2n+1)(2n+3)}=\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n(\frac{1}{2n+1}-\frac{1}{2n+3}).$$$$=\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{5}-\frac{1}{3})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{9}-\frac{1}{7})+...]$$$$=\frac{1}{2}[1+2(\frac{1}{5}+\frac{1}{9}+\frac{1}{13}+...)-2(\frac{1}{3}+\frac{1}{7}+\frac{1}{11}+...)]$$$$therestishistory.$$

Commented by abdo imad last updated on 18/Mar/18

$$letputS=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n+1)(2n+3)}and$$$$S\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{(2n+1)(2n+3)}wehaveS(-1)=Sbut$$$$S\left(x\right)=\frac{1}{2}\left(\sum _{n=0}^{\mathrm{\infty }}(\frac{1}{2n+1}-\frac{1}{2n+3})x^n\right)\Rightarrow forx\ne o$$$$2S\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{2n+1}-\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{2n+3}but$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{2n+3}=\frac{1}{3}+\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{2n+3}=\frac{1}{3}+\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n-1}}{2n+1}$$$$=\frac{1}{3}+\frac{1}{x}\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{2n+1}\Rightarrow$$$$2S\left(x\right)=-\frac{1}{3}+(1-\frac{1}{x})\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{2n+1}letcalculatefor-1⩽x<0$$$$A\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{2n+1}\Rightarrow A\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{(-x)}^n}{2n+1}$$$$=\frac{1}{\sqrt{-x}}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{\left(\sqrt{-x}\right)}^{2n+1}}{2n+1}=\frac{1}{\sqrt{-x}}\phi \left(\sqrt{-x}\right)with$$$$\phi \left(t\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{t}^{2n+1}}{2n+1}\Rightarrow {\phi }^{{}^{\prime }}\left(t\right)=\sum _{n=0}^{\mathrm{\infty }}{(-t^2)}^n=\frac{1}{1+t^2}$$$$\Rightarrow \phi \left(t\right)=arctant+\lambda but\lambda =\phi \left(0\right)=0\Rightarrow$$$$A\left(x\right)=\frac{1}{\sqrt{-x}}rctan\left(\sqrt{-x}\right)sofor-1⩽x<0wehave$$$$S\left(x\right)=-\frac{1}{6}+\frac{1}{2}\frac{x-1}{x}\frac{1}{\sqrt{-x}}arctan\left(\sqrt{-x}\right)$$$$=\frac{x-1}{2x\sqrt{-x}}arctan\left(\sqrt{-x}\right)-\frac{1}{6}and$$$$S(-1)=S=\frac{-2}{-2}arctan\left(1\right)-\frac{1}{6}=\frac{\pi }{4}-\frac{1}{6}.$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com