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Question Number 28814 by abdo imad last updated on 30/Jan/18

$$find\sum _{n=1}^{\mathrm{\infty }}\frac{1}{1^3+2^3+...+n^3}.$$

Commented by abdo imad last updated on 01/Feb/18

$$letput{S}_n=\sum _{k=1}^n\frac{1}{1^3+2^3+....+k^3}andS=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{1^3+2^3+...+n^3}={lim}_{n\to +\mathrm{\infty }}{S}_{n}weknowthat$$$$1^3+2^3+....k^3={\left(\frac{k(k+1)}{2}\right)}^2=\frac{k^2{(k+1)}^2}{4}so$$$$S_n=4\sum _{k=1}^n\frac{1}{k^2{(k+1)}^2}letdecomposeF\left(x\right)=\frac{1}{x^2{(x+1)}^2}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x+1}+\frac{d}{{(x+1)}^2}$$$$b={lim}_{x\to 0}{x}^{2}f\left(x\right)=1,d={lim}_{x\to -1}{(x+1)}^{2}F\left(x\right)=1so$$$$F\left(x\right)=\frac{a}{x}+\frac{1}{x^2}+\frac{c}{x+1}+\frac{1}{{(x+1)}^2}$$$$lim{}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+c\Rightarrow c=-aso$$$$F\left(x\right)=\frac{a}{x}-\frac{a}{x+1}+\frac{1}{x^2}+\frac{1}{{(x+1)}^2}$$$$F\left(1\right)=\frac{1}{4}=a-\frac{a}{2}++1+\frac{1}{4}\Rightarrow \frac{a}{2}+1=0\Rightarrow a=-2so$$$$F\left(x\right)=\frac{-2}{x}+\frac{2}{x+1}+\frac{1}{x^2}+\frac{1}{{(x+1)}^2}$$$$\sum _{k=1}^{n}F\left(k\right)=-2\sum _{k=1}^n\frac{1}{k}+2\sum _{k=1}^n\frac{1}{k+1}+\sum _{k=1}^n\frac{1}{k^2}$$$$+\sum _{k=1}^n\frac{1}{{(k+1)}^2}but$$$$\sum _{k=1}^n\frac{1}{k}=H_n$$$$\sum _{k=1}^n\frac{1}{k+1}=\sum _{k=2}^{n+1}\frac{1}{k}=H_{n+1}-1$$$$\sum _{k=1}^n\frac{1}{k^2}={\xi }_n\left(2\right)with{\xi }_n\left(x\right)=\sum _{k=1}^n\frac{1}{k^x}.$$$$\sum _{k=1}^n\frac{1}{{(k+1)}^2}=\sum _{k=2}^{n+1}\frac{1}{k^2}={\xi }_{n+1}\left(2\right)-1$$$$\sum _{k=1}^{n}F\left(k\right)=-2H_n+2H_{n+1}-2+{\xi }_n\left(2\right)+{\xi }_{n+1}\left(2\right)-1$$$$=2(H_{n+1}-H_n)+{\xi }_n\left(2\right)+{\xi }_{n+1}\left(2\right)-1$$$$\to 0+\frac{{\pi }^2}{6}+\frac{{\pi }^2}{6}-1=\frac{{\pi }^2}{3}-1$$$$and{lim}_{n\to +\mathrm{\infty }}{S}_n=4(\frac{{\pi }^2}{3}-1)=S.$$

Commented by abdo imad last updated on 01/Feb/18

$$\sum _{k=1}^{n}F\left(k\right)=2(H_{n+1}-H_n)+{\xi }_n\left(2\right)+{\xi }_{n+1}\left(2\right)-3$$$$\to 0+\frac{2{\pi }^2}{6}-3=\frac{{\pi }^2}{3}-3and$$$${lim}_{n\to +\mathrm{\infty }}{S}_n=4(\frac{{\pi }^2}{3}-3)=\frac{4{\pi }^2}{3}-12.$$

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