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Question Number 183376 by TUN last updated on 25/Dec/22

Commented by CElcedricjunior last updated on 25/Dec/22

$$\underset{\mathit{n}=1}{\sum ^{\mathrm{\infty }}}{\mathit{n}\mathit{x}}^{\mathit{n}}=\mathit{x}+2{\mathit{x}}^2+3{\mathit{x}}^3+...\mathrm{\infty }{\mathit{x}}^{\mathrm{\infty }}$$$$=\mathit{x}(1+2\mathit{x}+3{\mathit{x}}^2+....+\mathrm{\infty }{\mathit{x}}^{\mathrm{\infty }-1})$$$$\underset{x\to 0}{\lim }\frac{\underset{\mathit{n}=1}{\sum ^{\mathrm{\infty }}}{\mathit{n}\mathit{x}}^{\mathit{n}}}{\mathit{x}}=\underset{x\to 0}{\lim }(1+2x+...\mathrm{\infty }{\mathit{x}}^{\mathrm{\infty }-1})=1$$

Answered by TheSupreme last updated on 25/Dec/22

$$\frac{1}{x}\underset{n=1}{\sum ^{\mathrm{\infty }}}{nx}^n=\frac{1}{x}\underset{n=1}{\sum ^{\mathrm{\infty }}}D\left(x^{n+1}\right)=\frac{1}{x}\sum _{n=0}D\left(x^{n+2}\right)=$$$$=\frac{1}{x}D\left(x^2\mathrm{\Sigma }{x}^n\right)=\frac{1}{x}D\left(\frac{x^2}{1-x}\right)=\frac{1}{x}\left(\frac{2x(1-x)+x^2}{{(1-x)}^2}\right)=$$$$=\frac{1}{x}\frac{2x-x^2}{{(1-x)}^2}=\frac{2-x}{{(1-x)}^2}$$$$li\underset{x\to 0}{m}=2$$$$$$

Answered by Rasheed.Sindhi last updated on 25/Dec/22

$$\underset{x\to 0}{\lim }\frac{\underset{n=1}{\sum ^{\mathrm{\infty }}}{nx}^n}{x}$$$$\underset{x\to 0}{\lim }\left(\underset{n=1}{\sum ^{\mathrm{\infty }}}{nx}^{n-1}\right)$$$$\underset{x\to 0}{\lim }(1+2x+3x^2+4x^3+...)$$$$1+2x+3x^2+4x^3+...=S_{\mathrm{\infty }}$$$$x+2x^2+3x^3+4x...={xS}_{\mathrm{\infty }}$$$$S_{\mathrm{\infty }}-{xS}_{\mathrm{\infty }}=1+x+x^2+...=\frac{1}{1-x}$$$$S_{\mathrm{\infty }}=\frac{1}{{(1-x)}^2}$$$$\underset{x\to 0}{\lim }\left(\frac{1}{{(1-x)}^2}\right)=1$$

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