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$\sum_{n=1}^{\infty} nx^n$ for $x \neq 1$

It is quite obvious that for $q>1$ the sum will be $\infty$, but how to calculate it for $q<1$?

Also, here is a solution with a derivative, but I want to find one without the use of derivative.

$\sum_{n=1}^{\infty} nx^n = x+2x^2+3x^3+ \ldots=x(1+2x+3x^2+\ldots)=x(x'+(x^2)'+(x^3)'+\ldots)= x(x+x^2+x^3+\ldots)'=x \cdot \frac{1}{{(x-1)}^2}$

But as I said, I would like to find a solution without derivative.

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3 Answers 3

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$S = x + 2x^2 + 3x^3 + ...$

$xS = x^2 + 2x^3 + ...$

Subtracting

$S(1-x) = x + x^2 + x^3 + ...$

Using sum of infinite GP

$S(1-x) = \frac{x}{1-x}$

$S = \frac{x}{(1-x)^2}$

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hint: Consruct the sequence of partial sums and try to work out a closed formula then take the limit as $n \to \infty$. See the proof of the geometric series.

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Now that you have the answer, you can expand it in a Taylor series and see that it is right

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  • $\begingroup$ I don't want to use Taylor series, derivatives etc. The solution should be as basic as possible. $\endgroup$
    – Josh
    Dec 4, 2013 at 19:23

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