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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.


marked as duplicate by lab bhattacharjee, heropup, N. F. Taussig, Daniel W. Farlow, quid Feb 13 '16 at 14:05

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  • $\begingroup$ Binomial theorem uses summation to a finite number not infinity $\infty$- See 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. $\endgroup$ – NoChance Feb 13 '16 at 2:50
  • $\begingroup$ the generalized binomial theorem, for any value including non integer values of r, is an infinite sum, like a Taylor series $\endgroup$ – mentorship Feb 13 '16 at 2:54

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

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

  • $\begingroup$ would this be true without the qualification that $|x|<1$ ? $\endgroup$ – mentorship Feb 13 '16 at 3:02
  • 1
    $\begingroup$ @mentorship no. the series only holds for those restricted x values. $\endgroup$ – zz20s Feb 13 '16 at 3:02
  • $\begingroup$ @mentorship: No, then the sum diverges $\endgroup$ – Ross Millikan Feb 13 '16 at 3:03
  • $\begingroup$ I am wondering then if that is the closed form for this was not specified. $\endgroup$ – mentorship Feb 13 '16 at 3:03
  • $\begingroup$ Consider as an example, when $x=2$...1+2+4+8+16+32+64+...=-1/2? $\endgroup$ – Eleven-Eleven Feb 13 '16 at 3:03

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