# Power series

1. May 7, 2006

### kezman

Find the sum of the series:
$$\sum\limits_{n = 1}^\infty {nx^n }$$ if $$\left| x \right| < 1$$

I thought maybe with the geometric form, but im not sure.

Last edited: May 7, 2006
2. May 7, 2006

### Geekster

Is it asking for a number, or just if the series converges?

3. May 7, 2006

### kezman

I think for a general solution. It should converge.

4. May 7, 2006

### verd

Is it asking you to find a power series representation??

5. May 7, 2006

### d_leet

It already is a power series... It's asking for an expression for the sum.

6. May 7, 2006

### Geekster

You might start by writing out partial sums and see if that gets you anywhere.....

7. May 7, 2006

### shmoe

How does this differ from your usual geometric series?

8. May 7, 2006

### Curious3141

Hint : Call the original series S. Write out the first five or so terms in the series. Divide the series by x to get a new series (S/x). Now take the difference between this new series and the original series (S/x - S), term by term and see what you end up with.

The other way to do it is to differentiate a geometric series, but that's more complicated and unnecessary.

Last edited: May 7, 2006
9. May 8, 2006

### Pyrrhus

Last edited by a moderator: Apr 22, 2017
10. May 8, 2006

### Curious3141

Comparing series like these to derivatives of geometric series is a nice and interesting approach (I used to do this), but in most cases I've found that simply dividing or multiplying by x is an easier approach.

Last edited by a moderator: Apr 22, 2017
11. May 8, 2006

### shmoe

May as well have a third approach:

$$\sum_{n=1}^{\infty}nx^n=\sum_{n=1}^{\infty}\sum_{i=1}^{n}x^n$$

Change the order of summation (absolutely convergent series) then apply geometric series a couple of times. This is maybe the most complicated of the three, practice in rearranging summations never hurt.

Last edited: May 8, 2006
12. May 8, 2006

### kezman

thanks for all the hints.

The method I had to use is the derivative of the geometric series (similar to the one used for the maclaurin problem) using

$$\left( {\frac{1}{{1 - x}}} \right)^\prime = \sum\limits_{n = 0}^\infty {nx^{n - 1} }$$