I know that the Series $\sum_{n=1}^\infty n \cdot x^n $ converges to $\frac{x}{(x-1)^2}$ but I'm not sure how to show it. I'm pretty sure that has been asked before, but I wasn't able to find anything...
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1Must be one of the most often asked question on the site. – Did Jul 28 '13 at 09:39
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1You’ll find several solutions among the answers to this very slightly different question; at least one of them can be applied directly. – Brian M. Scott Jul 28 '13 at 09:43
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1Related : http://math.stackexchange.com/questions/366342/question-about-a-infinite-series and http://math.stackexchange.com/questions/67364/sequence-sum-question – lab bhattacharjee Jul 28 '13 at 09:57
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HINT:
Using this formula, if $|x|<1, \sum_{1\le r<\infty }x^r=\frac x{1-x}=-\frac1{x-1}-1$
Now, differentiate wrt $x$
lab bhattacharjee
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Let $$S_n=\sum_{k=1}^n kx^k$$
Assuming $|x|<1$, $$S_n-xS_n=\sum_{k=1}^n x^k-nx^{n+1}=\frac{x(1-x^n)}{1-x}-nx^{n+1}\Rightarrow S_n=\frac{x(1-x^n)}{(1-x)^2}-\frac{nx^{n+1}}{1-x}$$
Now, if $|x|<1$, then $$\limsup_{n\rightarrow \infty}\left|\frac{(n+1)x^{n+1}}{nx^n}\right|=|x|\limsup_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)=|x|<1$$So, by ratio test, the series $\displaystyle \sum_{k=1}^n kx^k$ converges and hence $$\lim_{n\rightarrow \infty}nx^{n+1}=x\lim_{n\rightarrow \infty}nx^{n}=0$$
So, $$S=\sum_{n=1}^{\infty}nx^n=\lim_{n\rightarrow \infty}S_n=\frac{x}{(1-x)^2}$$
Samrat Mukhopadhyay
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1Although this solution is now accepted, the line with an $\implies$ sign still has several mistakes. – Did Jul 28 '13 at 10:09
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@SamratMukhopadhyay, I think we still need to derive the value of $n\cdot x^{n+1}$ – lab bhattacharjee Jul 28 '13 at 10:09
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Oh, right, we have to show that $\lim_{n\rightarrow \infty}nx^{n+1}=0$, right? – Samrat Mukhopadhyay Jul 28 '13 at 10:15
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