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The following series converges when $|x| < 1$:

\begin{align*} \sum\limits_{n=1}^\infty n x^n \\ \end{align*}

Show that it converges to:

\begin{align*} \frac{x}{(1-x)^2} \\ \end{align*}

clay
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1 Answers1

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Consider $$x\cdot\frac{d}{dx}\sum_{n=1}^{\infty}x^n=x\cdot\frac{d}{dx}\frac{x}{1-x}\\=\frac{x}{(1-x)^2} \ \mbox{for} \ |x|<1$$ Now, in general, interchange of differentiation and infinite summation is not allowed, but for a power series, this is allowed within the radius of convergence. This means that for $|x|<1$ (which is indeed within the radius of convergence): $$x\cdot\frac{d}{dx}\sum_{n=1}^{\infty}x^n=x\cdot\sum_{n=1}^{\infty}\frac{d}{dx}x^n\\=x\cdot\sum_{n=1}^{\infty}nx^{n-1}\\ = \sum_{n=1}^{\infty}nx^{n}\\ =\frac{x}{(1-x)^2} \ \mbox{(from earlier)}$$

Jihoon Kang
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    Note that you're allowed to derive because the series converges uniformly in the radius of convergence. – Rab Aug 19 '17 at 00:10