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Question Number 167820 by MWSuSon last updated on 26/Mar/22

$$\mathrm{given}{\mathrm{x}}^2=\frac{{\pi }^2}{3}+4\mathrm{\Sigma }{(-1)}^{\mathrm{n}}\frac{\cos \left(\mathrm{nx}\right)}{{\mathrm{n}}^2},\mathrm{show}\mathrm{that}\mathrm{\Sigma }\frac{1}{{(2\mathrm{n}-1)}^2}=\frac{{\pi }^2}{8}$$

Answered by Mathspace last updated on 26/Mar/22

$$x^2=\frac{{\pi }^2}{3}+4\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}cos\left(nx\right)$$$$x=\frac{\pi }{2}\Rightarrow \frac{{\pi }^2}{4}=\frac{{\pi }^2}{3}+4\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{n^2}cos\left(\frac{n\pi }{2}\right)$$$$n=2m\Rightarrow cos\left(n\frac{\pi }{2}\right)={(-1)}^m$$$$n=2m+1\Rightarrow cos\left(\frac{n\pi }{2}\right)=0\Rightarrow$$$$4\sum _{m=1}^{\mathrm{\infty }}\frac{{(-1)}^{2m}{(-1)}^m}{4m^2}=\frac{{\pi }^2}{4}-\frac{{\pi }^2}{3}$$$$=-\frac{{\pi }^2}{12}\Rightarrow \sum _{m=1}^{\mathrm{\infty }}\frac{{(-1)}^m}{m^2}=-\frac{{\pi }^2}{12}\Rightarrow$$$$\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}-\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}=-\frac{{\pi }^2}{12}$$$$\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(2n+1)}^2}=\frac{1}{4}\frac{{\pi }^2}{6}+\frac{{\pi }^2}{12}$$$$=\frac{{\pi }^2+2{\pi }^2}{24}=\frac{3{\pi }^2}{24}=\frac{{\pi }^2}{8}$$$$n=r-1\Rightarrow \sum _{r=1}^{\mathrm{\infty }}\frac{1}{{(2r-1)}^2}=\frac{{\pi }^2}{8}$$

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