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Question Number 131428 by pticantor last updated on 04/Feb/21

$$\mathit{S}=\sum _{\mathit{n}⩾1}\frac{6}{\mathit{n}(\mathit{n}+1)(2\mathit{n}+1)}$$$$1)\mathit{s}\mathit{h}\mathit{o}\mathit{w}\mathit{t}\mathit{h}\mathit{a}\mathit{t}\mathit{S}\mathit{c}\mathit{o}\mathit{n}\mathit{v}\mathit{e}\mathit{r}\mathit{g}\mathit{e}\mathit{s}$$$$2)\mathit{g}\mathit{i}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{e}\mathit{v}\mathit{a}\mathit{l}\mathit{u}\mathit{e}\mathit{o}\mathit{f}\mathit{S}$$

Answered by JDamian last updated on 04/Feb/21

$$S=\underset{n⩾1}{\mathrm{\Sigma }}\frac{1}{\frac{\mathit{n}(\mathit{n}+1)(2\mathit{n}+1)}{6}}$$$$\left(1\right)$$$$a_n=\frac{\mathit{n}(\mathit{n}+1)(2\mathit{n}+1)}{6}=1^2+2^2+3^2+\cdot \cdot \cdot +n^2$$$$\frac{1}{a_n}=\frac{1}{1^2+2^2+\cdot \cdot \cdot +n^2}<\frac{1}{n^2}\mathrm{\forall }n>1$$$$\sum _{n⩾1}\frac{1}{a_n}<\sum _{n⩾1}\frac{1}{n^2}=\frac{{\pi }^2}{6}$$$$S<\frac{{\pi }^2}{6}$$

Answered by Dwaipayan Shikari last updated on 04/Feb/21

$$\sum _{n⩾1}\frac{6}{n(2n+1)}-\sum _{n⩾1}\frac{6}{(n+1)(2n+1)}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{6}{(n+1)(2n+3)}-(\underset{n⩾0}{\sum ^{\mathrm{\infty }}}\frac{6}{(n+1)(2n+1)}-\underset{n⩾0}{\sum ^1}\frac{6}{(n+1)(2n+1)})$$$$=6(\psi \left(\frac{3}{2}\right)+\psi \left(\frac{1}{2}\right))+6$$$$=6(2-4log\left(2\right))+6=6(3-4log\left(2\right))\psi \left(\frac{3}{2}\right)=2-2log\left(2\right)\psi \left(\frac{1}{2}\right)=-2log\left(2\right)$$$$$$

Answered by mathmax by abdo last updated on 04/Feb/21

$$1)\mathrm{we}\mathrm{have}\frac{6}{\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}\sim \frac{6}{{2\mathrm{n}}^3}=\frac{3}{{\mathrm{n}}^3}\mathrm{and}\mathrm{\Sigma }\frac{3}{{\mathrm{n}}^3}\mathrm{cv}\Rightarrow \mathrm{S}\mathrm{cv}$$$$2)\frac{\mathrm{S}}{6}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{w}}_{\mathrm{n}}\mathrm{with}{\mathrm{w}}_{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}(\mathrm{k}+1)(2\mathrm{k}+1)}$$$$\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}(\mathrm{x}+1)(2\mathrm{x}+1)},\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+1}+\frac{\mathrm{c}}{2\mathrm{x}+1}$$$$\mathrm{a}=1,\mathrm{b}=\frac{1}{(-1)(-1)}=1,\mathrm{c}=\frac{1}{(-\frac{1}{2})\frac{1}{2}}=-4\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}+1}-\frac{4}{2\mathrm{x}+1}$$$${\mathrm{w}}_{\mathrm{n}}=\sum _{{\mathrm{k}}^{}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}+\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+1}-4\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{2\mathrm{k}+1}\mathrm{but}$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}}\sim \log \left(\mathrm{n}\right)+\gamma$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{\mathrm{k}+1}=\sum _{\mathrm{k}=2}^{\mathrm{n}+1}\frac{1}{\mathrm{k}}={\mathrm{H}}_{\mathrm{n}+1}-1\sim \log (\mathrm{n}+1)+\gamma -1$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{2\mathrm{k}+1}=\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2\mathrm{n}+1}=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2\mathrm{n}}+\frac{1}{2\mathrm{n}+1}$$$$-1-\frac{1}{2}-\frac{1}{4}-....-\frac{1}{2\mathrm{n}}={\mathrm{H}}_{2\mathrm{n}+1}-\frac{1}{2}{\mathrm{H}}_{\mathrm{n}}-1\sim \log (2\mathrm{n}+1)+\gamma$$$$-\frac{1}{2}(\log \left(\mathrm{n}\right)+\gamma )-1=\log (2\mathrm{n}+1)-\frac{1}{2}\log \left(\mathrm{n}\right)+\frac{\gamma }{2}-1\Rightarrow$$$$-4\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{1}{2\mathrm{k}+1}\sim -4\log (2\mathrm{n}+1)+2\mathrm{logn}-2\gamma +4\Rightarrow$$$${\mathrm{W}}_{\mathrm{n}}\sim \log \left(\mathrm{n}\right)+\gamma +\log (\mathrm{n}+1)+\gamma -1-4\log (2\mathrm{n}+1)+2\log \left(\mathrm{n}\right)-2\gamma +4\Rightarrow$$$${\mathrm{W}}_{\mathrm{n}}\sim \log \left({\mathrm{n}}^3\right)+\log (\mathrm{n}+1)-4\log (2\mathrm{n}+1)+3\Rightarrow$$$${\mathrm{W}}_{\mathrm{n}}\sim \log \left(\frac{{\mathrm{n}}^4+{\mathrm{n}}^3}{{(2\mathrm{n}+1)}^4}\right)+3\Rightarrow {\mathrm{W}}_{\mathrm{n}}\to 3-4\log \left(2\right)\Rightarrow$$$$\mathrm{S}=6{\mathrm{W}}_{\mathrm{\infty }}=6(3-4\log \left(2\right))$$$$$$

Commented by pticantor last updated on 08/Feb/21

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{t}\mathit{o}\mathit{o}\mathit{m}\mathit{u}\mathit{c}\mathit{h}$$

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