Derive the sum of $\sum_{i=1}^n ix^{i-1}$

convergence-divergencepower seriessequences-and-seriessummation

For the series

$$1 + 2x + 3x^2 + 4x^3 + 5x^4 + … + nx^{n-1}+… $$

and $x \ne 1, |x| < 1$.

I need to find partial sums and finally, the sum $S_n$ of series.
Here is what I've tried:

  1. We can take a series $S_2 = 1 + x + x^2 + x^3 + x^4 + …$ so that
    $\frac{d(S_2)}{dx} = S_1$ (source series).
  2. For the $|x| < 1$ the sum of $S_2$ (here is geometric progression): $\frac{1-x^n}{1-x} = \frac{1}{1-x}$
  3. $S_1 = \frac{d(S_2)}{dx} = \frac{d(\frac{1}{1-x})}{dx} = \frac{1}{(1-x)^2}$

But this answer is incorrect. Where is my mistake? Thank you.

Best Answer

You've got a good notion!

Integrating the partial sum

$$1+2x+\cdots nx^{n-1}$$ gives you $$C+x+x^2+\cdots x^n,$$ for some constant $C,$ which is $$C-1+\frac{1-x^{n+1}}{1-x}.$$ Then, taking the derivative using the quotient rule gets you $$\begin{eqnarray}\frac{-(n+1)(1-x)x^n+1-x^{n+1}}{(1-x)^2} &=& \frac{-(n+1)x^n+(n+2)x^{n+1}+1-x^{n+1}}{(1-x)^2}\\ &=& \frac{nx^{n+1}-(n+1)x^n+1}{(1-x)^2}\end{eqnarray}$$ for your partial sum's closed form.

You've correctly found the closed form of the limit of the partial sums.

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