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

Question

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.

Answer 1

$$p_n(x):=\sum_{i=1}^n x^i$$ is a polynomial, which you can differentiate term-wise, giving the polynomial

$$p'_n(x):=\sum_{i=1}^n ix^{i-1}.$$

At the same time, $p(x)$ is the sum of terms of a geometric series, and for $x\ne1$,

$$p_n(x)=\frac{x^{n+1}-1}{x-1}-1.$$

Then, for all $x\ne1$,

$$p'_n(x)=\frac{(n+1)x^n}{x-1}-\frac{x^{n+1}-1}{(x-1)^2}.$$

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The limit exists for all $|x|<1$, and

$$p'_\infty(x)=\dfrac1{(x-1)^2}.$$

Answer 2

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.