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Question Number 130006 by Adel last updated on 21/Jan/21

Answered by Dwaipayan Shikari last updated on 21/Jan/21

$$S=x+2x^2+3x^3+....$$$$S(1-x)=x+x^2+x^3+...$$$$S=\frac{x}{{(1-x)}^2}x=\frac{1}{3}$$$$\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...=\frac{3}{4}$$

Commented by Adel last updated on 21/Jan/21

$$\mathrm{s}(1-\mathrm{x})=\mathrm{x}+{\mathrm{x}}^2+{\mathrm{x}}^3+\cdot \cdot \cdot \cdot$$$$\mathrm{is}\mathrm{way}?$$

Commented by Dwaipayan Shikari last updated on 21/Jan/21

$$S=x+2x^2+3x^3+...$$$$-xS=-x^2-2x^3-...$$$$S(1-x)=x+x^2+x^3+...(x+x^2+x^3+..=\frac{x}{1-x})$$

Commented by Adel last updated on 21/Jan/21

$$\mathrm{thanks}$$

Answered by mathmax by abdo last updated on 21/Jan/21

$$\mathrm{S}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\mathrm{n}}{3^{\mathrm{n}}}=\mathrm{s}\left(\frac{1}{3}\right)\mathrm{with}\mathrm{s}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}\mathrm{we}\mathrm{take}\mid \mathrm{x}\mid <1$$$$\mathrm{we}\mathrm{have}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{n}}=\frac{1}{1-\mathrm{x}}\mathrm{by}\mathrm{derivation}\mathrm{we}\mathrm{get}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}-1}=\frac{1}{{(1-\mathrm{x})}^2}$$$$\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{nx}}^{\mathrm{n}}=\frac{\mathrm{x}}{{(1-\mathrm{x})}^2}=\mathrm{s}\left(\mathrm{x}\right)\Rightarrow \mathrm{s}\left(\frac{1}{3}\right)=\frac{1}{3{(1-\frac{1}{3})}^2}$$$$=\frac{1}{3\times \frac{4}{9}}=\frac{3}{4}$$

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