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Question Number 46607 by maxmathsup by imad last updated on 29/Oct/18

$$calculate\sum _{(i,j)\in N^2}\frac{i+j}{3^{i+j}}$$

Commented by maxmathsup by imad last updated on 30/Oct/18

$$wehave{S}_n=\sum _{i=0}^{\mathrm{\infty }}\left(\sum _{j=0}^{\mathrm{\infty }}\frac{i+j}{3^{i+j}}\right)=\sum _{i=0}^{\mathrm{\infty }}{S}_i$$$$S_i=\frac{i}{3^i}\sum _{j=0}^{\mathrm{\infty }}\frac{1}{3^j}+\frac{1}{3^i}\sum _{j=0}^{\mathrm{\infty }}\frac{j}{3^j}but\sum _{j=0}^{\mathrm{\infty }}\frac{1}{3^j}=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}also$$$$\sum _{j=0}^{\mathrm{\infty }}\frac{j}{3^j}=\sum _{j=1}^{\mathrm{\infty }}j{\left(\frac{1}{3}\right)}^j=w\left(\frac{1}{3}\right)withw\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{nx}^n$$$$wehave\sum _{n=0}^{\mathrm{\infty }}{x}^n=\frac{1}{1-x}for\mid x\mid <1\Rightarrow \sum _{n=1}^{\mathrm{\infty }}{nx}^{n-1}=\frac{x}{{(1-x)}^2}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}{nx}^n=\frac{x^2}{{(1-x)}^2}\Rightarrow w\left(\frac{1}{3}\right)=\frac{1}{9{\left(\frac{2}{3}\right)}^2}=\frac{1}{4}\Rightarrow S_i=\frac{3}{2}\frac{i}{3^i}+\frac{1}{4.3^i}\Rightarrow$$$$\sum _{i=0}^{\mathrm{\infty }}{S}_i=\frac{3}{2}\sum _{i=0}^{\mathrm{\infty }}\frac{i}{3^i}+\frac{1}{4}\sum _{i=0}^{\mathrm{\infty }}\frac{1}{3^i}$$$$=\frac{3}{2}\frac{1}{4}+\frac{1}{4}\frac{3}{2}=\frac{3}{8}+\frac{3}{8}=\frac{6}{8}=\frac{3}{4}\Rightarrow S=\frac{3}{4}.$$$$$$

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