# Summation differentiation geometric series

## Homework Statement

find the sum for

$$\sum_{k=1}^{\infty} kx^{k}$$

## Homework Equations

$$\sum_{k=0}^{\infty} x^{k} = \frac{1}{1-x}; -1 < x < 1$$

## The Attempt at a Solution

$$\sum_{k=1}^{\infty} kx^{k} = \sum_{n=0}^{\infty}(n+1)x^{n+1} = x\sum_{n=0}^{\infty} (n+1)x^{n} = x \frac{d}{dx} \sum_{n=0}^{\infty}x^{n+1}$$

I'm not sure how to proceed; thoughts?

## Answers and Replies

rock.freak667
Homework Helper
I think from here

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

You should just differentiate both sides w.r.t. x and then change the index.

does it matter that it's xn+1 and not xn?

tiny-tim
Homework Helper
Hi Somefantastik!

(ooh, you just changed it! )

That's fine … now you need to get that ∑xn+1 to be part of a ∑xn starting at 0 (so you may need to subtract something ).

$$x \frac{d}{dx} \sum_{n=0}^{\infty}x^{n+1} = x \frac{d}{dx}\left[x\left(\sum_{n=0}^{\infty}x^{n}\right)\right] = x \frac{d}{dx}\left[x\frac{1}{1-x}\right]$$

tiny-tim