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Question Number 57647 by maxmathsup by imad last updated on 09/Apr/19

$$findapproximatevalueof\xi \left(3\right)byusingn-1⩽n⩽n+1fornintegr$$$$natural.$$

Commented by maxmathsup by imad last updated on 18/Apr/19

$$letdetermine{S}_2=\sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n-1)n^2}+1\Rightarrow S_2={lim}_{n\to +\mathrm{\infty }}\sum _{k=1}^n\frac{1}{k{(k+1)}^2}+1$$$$letdecomposeG\left(x\right)=\frac{1}{x{(x+1)}^2}\Rightarrow G\left(x\right)=\frac{a}{x}+\frac{b}{x+1}+\frac{c}{{(x+1)}^2}$$$$a={lim}_{x\to +\mathrm{\infty }}xG\left(x\right)=1,c={lim}_{x\to -1}{(x+1)}^{2}G\left(x\right)=-1\Rightarrow$$$$G\left(x\right)=\frac{1}{x}+\frac{b}{x+1}-\frac{1}{{(x+1)}^2}$$$$G\left(1\right)=\frac{1}{4}=1+\frac{b}{2}-\frac{1}{4}\Rightarrow 1=4+2b-1=3+2b\Rightarrow 2b=-2\Rightarrow b=-1\Rightarrow$$$$G\left(x\right)=\frac{1}{x}-\frac{1}{x+1}-\frac{1}{{(x+1)}^2}\Rightarrow \sum _{k=1}^{n}G\left(k\right)+1$$$$=\sum _{k=1}^n\frac{1}{k}-\sum _{k=1}^n\frac{1}{k+1}-\sum _{k=1}^n\frac{1}{{(k+1)}^2}+1$$$$=\sum _{k=1}^n\frac{1}{k}-\sum _{k=2}^{n+1}\frac{1}{k}-\sum _{k=2}^{n+1}\frac{1}{k^2}+1but$$$$\sum _{k=1}^n\frac{1}{k}=H_n,\sum _{k=2}^{n+1}\frac{1}{k}=H_{n+1}-1and\sum _{k=2}^{n+1}\frac{1}{k^2}={\xi }_{n+1}\left(2\right)-1\Rightarrow$$$$S_2=H_n-H_{n+1}+1-{\xi }_{n+1}\left(2\right)+1\to 2-\frac{{\pi }^2}{6}\Rightarrow \frac{{\pi }^2}{6}-\frac{1}{2}⩽\xi \left(3\right)⩽2-\frac{{\pi }^2}{6}$$$$\Rightarrow \frac{{\pi }^2-3}{6}⩽\xi \left(3\right)⩽\frac{12-{\pi }^2}{6}$$$$$$

Commented by maxmathsup by imad last updated on 18/Apr/19

$$wehave(n-1)n^2⩽n^3⩽(n+1)n^2\Rightarrow \frac{1}{(n+1)n^2}⩽\frac{1}{n^3}⩽\frac{1}{(n-1)n^2}\Rightarrow$$$$\sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n+1)n^2}⩽\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n^3}⩽\sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n-1)n^2}\Rightarrow \sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n+1)n^2}⩽\xi \left(3\right)-1$$$$⩽\sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n-1)n^2}\Rightarrow \sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n+1)n^2}+1⩽\xi \left(3\right)⩽\sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n-1)n^2}+1$$$$letdetermine{S}_1=1+\sum _{n=2}^{\mathrm{\infty }}\frac{1}{(n+1)n^2}\Rightarrow S_1=\frac{1}{2}+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n+1)n^2}$$$$S_1={lim}_{n\to +\mathrm{\infty }}{S}_{n}with{S}_n=\frac{1}{2}+\sum _{k=1}^n\frac{1}{(k+1)k^2}letdecomposeF\left(x\right)=\frac{1}{(x+1)x^2}$$$$F\left(x\right)=\frac{a}{x+1}+\frac{b}{x}+\frac{c}{x^2}$$$$a={lim}_{x\to -1}(x+1)F\left(x\right)=1,c={lim}_{x\to 0}{x}^{2}F\left(x\right)=1\Rightarrow F\left(x\right)=\frac{1}{x+1}+\frac{b}{x}+\frac{1}{x^2}$$$$F\left(1\right)=\frac{1}{2}=\frac{1}{2}+b+1\Rightarrow b=-1\Rightarrow F\left(x\right)=\frac{1}{x+1}-\frac{1}{x}+\frac{1}{x^2}\Rightarrow$$$$\sum _{k=1}^{n}F\left(k\right)=\sum _{k=1}^n\frac{1}{k+1}-\sum _{k=1}^n\frac{1}{k}+\sum _{k=1}^n\frac{1}{k^2}$$$$=\sum _{k=2}^{n+1}\frac{1}{k}-\sum _{k=1}^n\frac{1}{k}+\sum _{k=1}^n\frac{1}{k^2}=-1-\frac{1}{n+1}+\sum _{k=1}^n\frac{1}{k^2}\to \frac{{\pi }^2}{6}-1\Rightarrow$$$$S_1=\frac{{\pi }^2}{6}-\frac{1}{2}$$

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