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Question Number 29505 by abdo imad last updated on 09/Feb/18

$$letgive{u}_n=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{{(k+1)}^2{2}^k}find{lim}_{n\to \mathrm{\infty }}{u}_n.$$

Commented by abdo imad last updated on 11/Feb/18

$$u_n=\sum _{k=0}^n\frac{1}{{(k+1)}^2{2}^k}.$$

Commented by abdo imad last updated on 13/Feb/18

$$wehave{u}_n=\sum _{k=1}^{n+1}\frac{1}{k^2{2}^{k-1}}=2\sum _{k=1}^{n+1}\frac{1}{k^2}{\left(\frac{1}{2}\right)}^{k}butthserie$$$$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}{\left(\frac{1}{2}\right)}^{k}isconvrgenteletputw\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n^2}$$$$for\mid x\mid <1wehave{w}^{{}^{\prime }}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^{n-1}}{n}\Rightarrow {xw}^{{}^{\prime }}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}$$$$\Rightarrow w^{{}^{\prime }}\left(x\right)+{xw}^{{}^{″}}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{x}^{n-1}=\sum _{n=0}^{\mathrm{\infty }}{x}^n=\frac{1}{1-x}sowis$$$$solutionforthed.e{y}^{{}^{\prime }}+{xy}^{{}^{″}}=\frac{1}{1-x}for{y}^{{}^{\prime }}=z\Rightarrow$$$$z+{xz}^{{}^{\prime }}=\frac{1}{1-x}h.e\Rightarrow z+{xz}^{{}^{\prime }}=0\Rightarrow {xz}^{{}^{\prime }}=-z\Rightarrow \frac{z^{{}^{\prime }}}{z}=\frac{-1}{x}$$$$\Rightarrow ln\mid z\mid =-ln\mid x\mid +k\Rightarrow z=\frac{\lambda }{x}for0<x<1mvc\Rightarrow$$$$z^{{}^{\prime }}=\frac{{\lambda }^{{}^{\prime }}}{x}-\frac{\lambda }{x^2}\Rightarrow \frac{\lambda }{x}+{\lambda }^{{}^{\prime }}-\frac{\lambda }{x}=\frac{1}{1-x}\Rightarrow {\lambda }^{{}^{\prime }}=\frac{1}{1-x}\Rightarrow$$$$\lambda \left(x\right)=-ln(1-x)+kandk=\lambda \left(0\right)=0$$$$z\left(x\right)=-\frac{ln(1-x)}{x}=y^{{}^{\prime }}\Rightarrow y\left(x\right)=-{\int }_0^x\frac{ln(1-t)}{t}dt+c$$$$c=y\left(0\right)=0\Rightarrow w\left(x\right)=-{\int }_0^x\frac{ln(1-t)}{t}dtand$$$$S=2w\left(\frac{1}{2}\right)=-2{\int }_0^{\frac{1}{2}}\frac{ln(1-t)}{t}dt.....becontinued....$$

Commented by abdo imad last updated on 14/Feb/18

$$wehave{lim}_{n\to \mathrm{\infty }}{u}_n=2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}{\left(\frac{1}{2}\right)}^{n}letput$$$$S\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{x}^n=\frac{1}{1-x}\Rightarrow {\int }_0^{x}S\left(t\right)dt=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{n+1}}{n+1}=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}$$$$=-ln(1-x)for0<x<1(\lambda =0)\Rightarrow$$$$x\sum _{n=1}^{\mathrm{\infty }}\frac{x^{n-1}}{n}=-ln(1-x)\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{x^{n-1}}{n}=-\frac{ln(1-x)}{x}\Rightarrow$$$${\int }_0^x\left(\sum _{n=1}^{\mathrm{\infty }}\frac{t^{n-1}}{n}\right)dt=-{\int }_0^x\frac{ln(1-t)}{t}+\lambda \Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n^2}=-{\int }_0^x\frac{ln(1-t)}{t}(\lambda =0)and$$$$2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}{\left(\frac{1}{2}\right)}^n=-{\int }_0^{\frac{1}{2}}\frac{ln(1-t)}{t}dtch.t=cos\theta$$$${\int }_0^{\frac{1}{2}}(...)dt={\int }_{\frac{\pi }{2}}^{\frac{\pi }{3}}\frac{ln(1-cos\theta )}{cos\theta }sin\theta d\theta$$$$={\int }_{\frac{\pi }{2}}^{\frac{\pi }{3}}tan\theta ln(1-cos\theta )d\theta ....becontinued...$$

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