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Question Number 62556 by naka3546 last updated on 22/Jun/19

$$\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+...=\frac{a}{b},a,b\in {\mathbb{Z}}^{+}$$

Commented by Prithwish sen last updated on 22/Jun/19

$${\mathrm{t}}_{\mathrm{n}}=\frac{1}{3^{\mathrm{n}}}+\frac{2\mathrm{n}}{3^{\mathrm{n}}}+\frac{{\mathrm{n}}^2}{3^{\mathrm{n}}}$$$${\mathrm{t}}_1=\frac{1}{3}+2\frac{1}{3}+\frac{1^2}{3}$$$${\mathrm{t}}_2=\frac{1}{3^2}+2\frac{2}{3^2}+\frac{2^2}{3^2}$$$$............................$$$$\mathrm{S}=(\frac{1}{3}+\frac{1}{3^2}+...)+2(\frac{1}{3}+\frac{2}{3^2}+.........)+(\frac{1^2}{3}+\frac{2^2}{3^2}+\frac{3^2}{3^3}+......)$$$$={\mathrm{I}}_1+{\mathrm{I}}_2+{\mathrm{I}}_3$$$$\mathrm{Now}$$$${\mathrm{I}}_1=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}$$$$\mathrm{For}{\mathrm{I}}_2$$$$\frac{{\mathrm{t}}_{\mathrm{n}+1}}{{\mathrm{t}}_{\mathrm{n}}}=\frac{\mathrm{n}+1}{3\mathrm{n}}\to \frac{1}{3}<1\mathrm{as}\mathrm{n}\to \mathrm{\infty }$$$$\therefore {\mathrm{I}}_2\mathrm{converges}$$$$\mathrm{For}{\mathrm{I}}_3$$$$\frac{{\mathrm{t}}_{\mathrm{n}+1}}{{\mathrm{t}}_{\mathrm{n}}}=\frac{1}{3}.\frac{{(\mathrm{n}+1)}^2}{{\mathrm{n}}^2}=\frac{1}{3}.(1+\frac{2}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2})\to \frac{1}{3}\mathrm{as}\mathrm{n}\to \mathrm{\infty }$$$$\mathrm{thus}{\mathrm{I}}_3\mathrm{also}\mathrm{converges}$$$$\therefore \mathrm{the}\mathrm{entire}\mathrm{series}\mathrm{is}\mathrm{convergent}$$$$\mathrm{Is}\mathrm{it}\mathrm{ok}?$$$$\mathrm{Waiting}\mathrm{for}\mathrm{your}\mathrm{feedback}.$$$$$$$$$$$$$$

Commented by mathmax by abdo last updated on 22/Jun/19

$$S=\sum _{n=1}^{\mathrm{\infty }}\frac{\sum _{k=0}^n(2k+1)}{3^n}letthesequence{u}_k=2k+1issrithmeticwith{u}_0=1$$$$andr=2\Rightarrow \sum _{k=0}^n{u}_k=u_0+u_1+...+u_n=\frac{n+1}{2}(u_o+u_n)=\frac{n+1}{2}(1+2n+1)$$$$={(n+1)}^2\Rightarrow S=\sum _{n=1}^{\mathrm{\infty }}\frac{{(n+1)}^2}{3^n}=\sum _{n=2}^{\mathrm{\infty }}\frac{n^2}{3^{n-1}}=3\sum _{n=2}^{\mathrm{\infty }}\frac{n^2}{3^n}$$$$\sum _{n=2}^{\mathrm{\infty }}\frac{n^2}{3^n}=\sum _{n=0}^{\mathrm{\infty }}\frac{n^2}{3^n}-\frac{1}{3}letw\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{x}^{n}with\mid x\mid <1wehave$$$$w^{{}^{\prime }}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{nx}^{n-1}\Rightarrow {xw}^{{}^{\prime }}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{nx}^n\Rightarrow w^{{}^{\prime }}\left(x\right)+x{w}^{\left(2\right)}\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{n}^2{x}^{n-1}\Rightarrow$$$$x\{w^{{}^{\prime }}\left(x\right)+x{w}^{\left(2\right)}\left(x\right)\}=\sum _{n=1}^{\mathrm{\infty }}{n}^2{x}^{n}butw\left(x\right)=\frac{1}{1-x}\Rightarrow w^{{}^{\prime }}\left(x\right)=\frac{1}{{(1-x)}^2}\Rightarrow$$$$w^{\left(2\right)}\left(x\right)=-\frac{2(-1)(1-x)}{{(1-x)}^4}=\frac{2}{{(1-x)}^3}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}{n}^2{x}^n=x\{\frac{1}{{(1-x)}^2}+\frac{2x}{{(1-x)}^3}\}=\frac{x}{{(1-x)}^2}+\frac{2x^2}{{(1-x)}^3}=\frac{x(1-x)+2x^2}{{(1-x)}^3}=\frac{x^2+x}{{(1-x)}^3}$$$$and\sum _{n=1}^{\mathrm{\infty }}\frac{n^2}{3^n}=w\left(\frac{1}{3}\right)=\frac{\frac{1}{9}+\frac{1}{3}}{{\left(\frac{2}{3}\right)}^3}=\frac{4}{8}=\frac{1}{2}\Rightarrow S=3\{\frac{1}{2}-\frac{1}{3}\}=\frac{3}{2}-1=\frac{1}{2}$$$$$$

Answered by Smail last updated on 23/Jun/19

$$\frac{a}{b}=\frac{1+3}{3}+\frac{1+3+5}{3^2}+...$$$$Knowingthat1+3+5+7+...+2k-1=k^2$$$$So,\frac{a}{b}=\frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+...$$$$\frac{a}{b}=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{n^2}{3^{n-1}}$$$$Knowingthat\underset{n=0}{\sum ^{\mathrm{\infty }}}{x}^n=\frac{1}{1-x}with\mid x\mid <1$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{nx}^n=\frac{x}{{(1-x)}^2}$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{n}^2{x}^{n-1}=\frac{(1-x)+2x}{{(1-x)}^3}=\frac{1+x}{{(1-x)}^3}$$$$\underset{n=2}{\sum ^{\mathrm{\infty }}}{n}^2{x}^{n-1}+1=\frac{1+x}{{(1-x)}^3}$$$$\underset{n=2}{\sum ^{\mathrm{\infty }}}{n}^2{x}^{n-1}=\frac{1+x}{{(1-x)}^3}-1$$$$Forx=\frac{1}{3}$$$$\frac{a}{b}=\underset{n=2}{\sum ^{\mathrm{\infty }}}\frac{n^2}{3^{n-1}}=\frac{1+\frac{1}{3}}{{(1-\frac{1}{3})}^3}-1=\frac{7}{2}$$$$Thus,a=7andb=2$$

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