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Question Number 38518 by math khazana by abdo last updated on 26/Jun/18

$$calculate\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2{(2n-1)}^2}$$

Commented by prof Abdo imad last updated on 27/Jun/18

$$letput{S}_n=\sum _{k=1}^n\frac{1}{k^2{(2k-1)}^2}andletdecompose$$$$F\left(x\right)=\frac{1}{x^2{(2x-1)}^2}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{2x-1}+\frac{d}{{(2x-1)}^2}$$$$b={lim}_{x\to 0}{x}^{2}F\left(x\right)=1$$$$d={lim}_{x\to \frac{1}{2}}{(2x-1)}^{2}F\left(x\right)=4\Rightarrow$$$$F\left(x\right)=\frac{a}{x}+\frac{1}{x^2}+\frac{c}{2x-1}+\frac{4}{{(2x-1)}^2}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+\frac{c}{2}\Rightarrow 2a+c=0\Rightarrow c=-2a$$$$F\left(x\right)=\frac{a}{x}-\frac{2a}{2x-1}+\frac{1}{x^2}+\frac{4}{{(2x-1)}^2}$$$$F\left(1\right)=1=a-2a+1+4\Rightarrow -a+4=0\Rightarrow a=4$$$$F\left(x\right)=\frac{4}{x}-\frac{8}{2x-1}+\frac{1}{x^2}+\frac{4}{{(2x-1)}^2}$$$$S_n=\sum _{k=1}^{n}F\left(k\right)=4\sum _{k=1}^n\frac{1}{k}-8\sum _{k=1}^n\frac{1}{2k-1}$$$$+\sum _{k=1}^n\frac{1}{k^2}+4\sum _{k=1}^n\frac{1}{{(2k-1)}^2}but$$$$\sum _{k=1}^n\frac{1}{k}=H_n$$$$\sum _{k=1}^n\frac{1}{2k-1}=1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}$$$$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2n-1}+\frac{1}{2n}$$$$-\frac{1}{2}{H}_n=H_{2n}-\frac{1}{2}{H}_n$$$$\sum _{k=1}^n\frac{1}{k^2}={\xi }_n\left(2\right)$$$$\sum _{k=1}^n\frac{1}{{(2k-1)}^2}=\sum _{k=2}^{n+1}\frac{1}{{(2k+1)}^2}$$$$=\sum _{k=0}^{n+1}\frac{1}{{(2k+1)}^2}\Rightarrow$$$$S_n=4H_n-8H_{2n}+4H_n+{\xi }_n\left(2\right)+4\sum _{k=0}^{n+1}\frac{1}{{(2k+1)}^2}$$$$S_n=-8(H_{2n}-H_n)+{\xi }_n\left(2\right)+4\sum _{k=0}^{n+1}\frac{1}{{(2k+1)}^2}$$$${lim}_{n\to +\mathrm{\infty }}{S}_n=-8ln\left(2\right)+\frac{{\pi }^2}{6}+4.\frac{{\pi }^2}{8}$$$$=-8ln\left(2\right)+\frac{{\pi }^2}{6}+\frac{{\pi }^2}{2}=\frac{4{\pi }^2}{6}-8ln\left(2\right)$$$$=\frac{2{\pi }^2}{3}-8ln\left(2\right).$$

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