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Question Number 145996 by iloveisrael last updated on 10/Jul/21

$$\mathrm{find}\mathrm{the}\mathrm{sum}\mathrm{of}$$$$\mathrm{x}-{2\mathrm{x}}^2+{3\mathrm{x}}^3-{4\mathrm{x}}^4+{5\mathrm{x}}^5-{6\mathrm{x}}^6+...$$$$\mathrm{where}\mid \mathrm{x}\mid <1$$

Answered by mathmax by abdo last updated on 10/Jul/21

$$\mathrm{S}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}\Rightarrow \mathrm{S}\left(\mathrm{x}\right)=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{(-\mathrm{x})}^{\mathrm{n}}$$$$\mathrm{we}\mathrm{have}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-\mathrm{x})}^{\mathrm{n}}=\frac{1}{1+\mathrm{x}}\mathrm{by}\mathrm{derivation}\mathrm{we}\mathrm{get}$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{nx}}^{\mathrm{n}-1}=-\frac{1}{{(1+\mathrm{x})}^2}\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}\mathrm{n}{(-\mathrm{x})}^{\mathrm{n}}=\frac{-\mathrm{x}}{{(1+\mathrm{x})}^2}\Rightarrow$$$$\mathrm{S}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{{(1+\mathrm{x})}^2}$$

Answered by mr W last updated on 10/Jul/21

$$S=\mathrm{x}-{2\mathrm{x}}^2+{3\mathrm{x}}^3-{4\mathrm{x}}^4+{5\mathrm{x}}^5-{6\mathrm{x}}^6+...$$$$xS={\mathrm{x}}^2-{2\mathrm{x}}^3+{3\mathrm{x}}^4-{4\mathrm{x}}^5+{5\mathrm{x}}^6+...$$$$(1+x)S=x-x^2+x^3-x^4+x^5-x^6+...$$$$(1+x)S=\frac{x}{1+x}$$$$\Rightarrow S=\frac{x}{{(1+x)}^2}$$

Answered by qaz last updated on 10/Jul/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}-1}{\mathrm{nx}}^{\mathrm{n}}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}(\mathrm{n}+1){\mathrm{x}}^{\mathrm{n}+1}$$$$=\mathrm{x}(\mathrm{yD}+1){\mid }_{\mathrm{y}=-\mathrm{x}}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{y}}^{\mathrm{n}}$$$$=\mathrm{x}(\mathrm{yD}+1){\mid }_{\mathrm{y}=-\mathrm{x}}\frac{1}{1-\mathrm{y}}$$$$=\mathrm{x}{[\frac{\mathrm{y}}{{(1-\mathrm{y})}^2}+\frac{1}{1-\mathrm{y}}]}_{\mathrm{y}=-\mathrm{x}}$$$$=\frac{\mathrm{x}}{{(1+\mathrm{x})}^2}$$

Commented by qaz last updated on 10/Jul/21

$$\mathrm{D}=\frac{\mathrm{d}}{\mathrm{dx}}$$$$\mathrm{Dx}-\mathrm{xD}=1$$$$\mathrm{xDxD}\cdot ...\cdot \mathrm{xD}={\left(\mathrm{xD}\right)}^{\mathrm{n}}={(\mathrm{Dx}-1)}^{\mathrm{n}}=\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}\left(\begin{array}{c}\mathrm{n}\\ \mathrm{k}\end{array}\right){\left(\mathrm{Dx}\right)}^{\mathrm{k}}{(-1)}^{\mathrm{n}-\mathrm{k}}$$$${\mathrm{D}}^{\mathrm{n}}\mathrm{x}-{\mathrm{xD}}^{\mathrm{n}}={\mathrm{nD}}^{\mathrm{n}-1}$$

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