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Question Number 33588 by abdo imad last updated on 19/Apr/18

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

Commented by abdo imad last updated on 20/Apr/18

$$letput{S}_n=\sum _{k=1}^n\frac{1}{k^2(k+1)(2k+1)}andletdecompose$$$$F\left(x\right)=\frac{1}{x^2(x+1)(2x+1)}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x+1}+\frac{d}{2x+1}$$$$b={lim}_{x\to 0}{x}^{2}F\left(x\right)=1$$$$c={lim}_{x\to -1}(x+1)F\left(x\right)=\frac{1}{-1}=-1$$$$d={lim}_{x\to -\frac{1}{2}}(2x+1)F\left(x\right)=\frac{1}{\frac{1}{4}.\frac{1}{2}}=8\Rightarrow$$$$F\left(x\right)=\frac{a}{x}+\frac{1}{x^2}-\frac{1}{x+1}+\frac{8}{2x+1}$$$$F\left(1\right)=\frac{1}{6}=a+1-\frac{1}{2}+\frac{8}{3}=a+\frac{1}{2}+\frac{8}{3}\Rightarrow$$$$1=6a+3+16=6a+19\Rightarrow 6a=-18\Rightarrow a=-3\Rightarrow$$$$F\left(x\right)=-\frac{3}{x}+\frac{1}{x^2}-\frac{1}{x+1}+\frac{8}{2x+1}$$$$S_n=-3\sum _{k=1}^n\frac{1}{k}+\sum _{k=1}^n\frac{1}{k^2}-\sum _{k=1}^n\frac{1}{k+1}+8\sum _{k=1}^n\frac{1}{2k+1}$$$$\sum _{k=1}^n\frac{1}{k}=H_n$$$$\sum _{k=1}^n\frac{1}{k^2}={\xi }_n\left(2\right)({\xi }_n\left(x\right)=\sum _{k=1}^n\frac{1}{k^x})$$$$\sum _{k=1}^n\frac{1}{k+1}=\sum _{k=2}^{n+1}\frac{1}{k}=H_{n+1}-1$$$$\sum _{k=1}^n\frac{1}{2k+1}=\frac{1}{3}+\frac{1}{5}+......+\frac{1}{2n+1}$$$$=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2n}+\frac{1}{2n+1}-1-\frac{1}{2}(1+\frac{1}{2}+...+\frac{1}{n})$$$$=H_{2n+1}-\frac{1}{2}{H}_n-1\Rightarrow$$$$S_n=-3H_n+{\xi }_n\left(2\right)-H_{n+1}+1+8H_{2n+1}-4H_n-8$$$$S_n=-7H_n-H_{n+1}+8H_{2n+1}-7+{\xi }_n\left(2\right)\Rightarrow$$$$S_n=-7(ln\left(n\right)+\gamma +o\left(\frac{1}{n}\right))-(ln(n+1)+\gamma +o\left(\frac{1}{n}\right))$$$$+8(ln(2n+1)+\gamma +o\left(\frac{1}{n}\right))+{\xi }_n\left(2\right)-7$$$$=-ln\left(n^7(n+1)\right)+ln\left({(2n+1)}^8\right)+{\xi }_n\left(2\right)-7$$$$=ln\left(\frac{{(2n+1)}^8}{n^8+n^7}\right)+{\xi }_n\left(2\right)-7\Rightarrow$$$${lim}_{n\to +\mathrm{\infty }}{S}_n=ln\left(2^8\right)+\xi \left(2\right)-7=8ln\left(2\right)+\frac{{\pi }^2}{6}-7.so$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2(n+1)(2n+1)}=8ln\left(2\right)+\frac{{\pi }^2}{6}-7.$$$$$$

Answered by sma3l2996 last updated on 19/Apr/18

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2(n+1)(2n+1)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{a}{n}+\frac{b}{n^2}+\frac{c}{n+1}+\frac{d}{2n+1})$$$$witha=-3;b=1;c=-1;d=8$$$$so\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2(n+1)(2n+1)}=\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{-3}{n}+\frac{1}{n^2}-\frac{1}{n+1}+\frac{8}{2n+1})$$$${let}^{\prime }scalculateA=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{8}{2n+1}$$$$A=8(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...)=8[(\frac{1}{3}+\frac{1}{5}+...)+(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...)-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...)]$$$$=8[\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{1}{2}(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...)]$$$$=8(\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{1}{2}\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-\frac{1}{2})=4\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-4$$$$so\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{8}{2n+1}-\frac{3}{n}-\frac{1}{n+1})=4\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-4-3(\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n}+\frac{1}{1})-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n+1}$$$$letk=n+1$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}(\frac{8}{2n+1}-\frac{3}{n}-\frac{1}{n+1})=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{1}{n}-7-\underset{k=2}{\sum ^{\mathrm{\infty }}}\frac{1}{k}=-7$$$$weknowthat\zeta \left(2\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2}=\frac{{\pi }^2}{6}\left(\zeta \left(x\right)isRiemannzetafunction\right)$$$$Asresults$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^2(n+1)(2n+1)}=\frac{{\pi }^2}{6}-7$$

Commented by abdo imad last updated on 20/Apr/18

$$sirsma3lyouhavecommitedaerrorofcalculusbecause$$$$theserieispositifbutthevalue\frac{{\pi }^2}{6}-7isnegativebut$$$$nevermindyourmethodiscorrect...$$

Commented by sma3l2996 last updated on 20/Apr/18

$$ThatwhatIjustsaw.Thereisanerrorinmywork.$$$$Canyoucheckmyworkplease?$$$$$$

Commented by abdo imad last updated on 20/Apr/18

$$iadviseyousirsma3ltousetheharmonicsequence{H}_n$$$$tookalookinmymethod....$$

Commented by sma3l2996 last updated on 20/Apr/18

$$Okthankyou$$

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