# Summation differentiation geometric series

1. Feb 2, 2010

### Somefantastik

1. The problem statement, all variables and given/known data
find the sum for

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

2. Relevant equations

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

3. 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?

2. Feb 2, 2010

### rock.freak667

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.

3. Feb 2, 2010

### Somefantastik

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

4. Feb 2, 2010

### tiny-tim

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 ).

5. Feb 2, 2010

### Somefantastik

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

6. Feb 2, 2010

### tiny-tim

Yes, that's fine also.

( x/(1-x) = 1/(1-x) - 1 )

(though in the form 1/(1-x) - 1, it's easier to differentiate! )