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Find a closed formula for $\sum_{n=1}^\infty nx^{n-1}$

I am trying to use the derivative of generalized binomial theorem, $\frac{d}{dx}[(x+1)^r=\sum_{n=0}^\infty \binom{r}{n}x^n] =r(x+1)^{r-1}=\sum_{n=1}^\infty \binom{r}{n}nx^{n-1}$

However, I am not sure how to "get rid" of the binomial term.

meariMD
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  • Binomial theorem uses summation to a finite number not infinity $\infty$- See https://en.wikipedia.org/wiki/Binomial_theorem. Also for such infinite sum to converge you need to put a condition on $x$, otherwise the series will not converge. – NoChance Feb 13 '16 at 02:50
  • the generalized binomial theorem, for any value including non integer values of r, is an infinite sum, like a Taylor series – meariMD Feb 13 '16 at 02:54

1 Answers1

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Note that for $|x|<1$, we have $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$

Taking derivatives of both sides

$$\sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$

On the LHS, when $n=0$, the term vanishes. Thus

$$\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$

Eleven-Eleven
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