Closed form for $\sum^{\infty}_{{i=n}}ix^{i-1}$

Question

How can I find a closed form for:

$$\sum^{\infty}_{{i=n}}ix^{i-1}$$

It looks like that's something to do with the derivative

Answer 1

$$\sum^\infty_{r=n}rx^{r-1}=\dfrac {d\left(\sum^\infty_{r=n}x^r\right)}{dx}$$

From here, $\sum^\infty_{r=n}x^r=\dfrac{x^n}{1-x}$ if $|x|<1$

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Alternatively,

let $S_n=\sum^m_{r=n}rx^{r-1}=n\cdot x^{n-1}+(n+1)x^n+(n+2)x^{n+1}+\cdots+(m-1)x^{m-2}+mx^{m-1}$

So, multiplying by $x,$ $x\cdot S_n=\sum^\infty_{r=n}rx^r=n x^n+(n+1)x^{n+1}+(n+2)x^{n+2}+\cdots+(m-1)x^{m-1}+mx^m$

On subtraction, $$(1-x)S_n=nx^{n-1}+(x^n+x^{n+1}+\cdots+x^{m-1})-mx^m$$ $$=nx^{n-1}+x^n\frac{(1-x^{m-n})}{1-x}-mx^m=nx^{n-1}+\frac{(x^n-x^m)}{1-x}-mx^m$$

Now, if $m\to\infty$ and $|x|<1,x^m\to0$

From this, $\lim_{m\to\infty}mx^m=0$ if $|x|<1$

Answer 2

Recall the power series:

$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$

holds for all $|x|<1$.

Then take the derivates, you get:

$$\frac{1}{(1-x)^2}=\sum_{n\geq1}nx^{n-1}$$