# Help with infinite sum.

1. Nov 14, 2011

### Beer-monster

Could someone suggest how I could go about calculating this infinite sum?

$$\sum_{n=1}^{\infty}n^{2}x^{n}$$

I'm having some trouble as this series does not have a constant ratio, difference or matches any standard series I can think of at the moment.

Thanks

2. Nov 14, 2011

### Dick

You derive something like that by looking at derivatives of the geometric series. You can sum x^n, n=0 to infinity, right? Take a derivative of both sides. What's sum n*x^n, k=0 to infinity? Now take another derivative.

3. Nov 14, 2011

### Beer-monster

So, take the derivative, do the sum then take the new derivative?

So the first sum would be the derivative of $nx^{n-1}$

Or just look at the what derivatives I can solve that give me $n^{2}x^{n}$?

4. Nov 14, 2011

### Dick

So did you get the sum of n*x^(n-1)? What is it?

Last edited: Nov 14, 2011
5. Nov 14, 2011

### Beer-monster

It's the sum of increasing derivatives of $x^{n}$.

i.e. $$\frac{d1}{dx}+\frac{dx}{dx}+\frac{d^{2}x^{2}}{dx^{2}} ...\frac{d^{n}x^{n}}{dx^{n}}$$

6. Nov 14, 2011

### Dick

No, that's not what I mean and it's not right. sum x^n=1/(1-x), right? Differentiate BOTH sides of the equation.

7. Nov 14, 2011

### Beer-monster

As so:

$$\frac{d}{dx}\sum{x^{n}}=\sum \frac{d}{dx}x^n = \frac{d}{dx} \frac{1}{1-x} = \frac{1}{(1-x)^{2}}$$

8. Nov 14, 2011

### Dick

Ok, so sum n=0 to infinity n*x^(n-1)=1/(1-x)^2. Right? Now take a second derivative.

9. Nov 14, 2011

### Beer-monster

You mean

$$\sum n(n-1)x^{n-2} = \frac{-2}{(1-x)^{3}}$$

Then do I remove the extra factor of x^{n-2} by adding on both sides to get my final answer?

10. Nov 15, 2011

### Dick

You've got a sign error in that derivative. If you know sum n(n-1)x^(n-2) then you can find sum n(n-1)x^n, right? They differ by a factor of x^2. Now n(n-1)*x^n=n^2*x^n-n*x^n. That's the general strategy. You'll need to pay some attention to the details of the limits to get the right answer.

11. Nov 15, 2011

### Beer-monster

I think I get the idea, but I'm getting something messy out.

$$\frac{2x^{2}}{(1-x)^{3}}+\frac{1}{(1-x)}$$

To take care of the fact that the sum is between 1 and infinity not 0 and infinity can I just subtract the n=0 term from my calculated sum?

12. Nov 15, 2011

### Ray Vickson

Well, just look at the n=0 term in sum n*x^(n-1), etc. You really can answer your questions for yourself.

RGV

13. Nov 15, 2011

### Beer-monster

Yes. I realise that came out a bit more stupid than I intended it to. My apologies for not being specific enough.

However, I was mostly thinking in general terms. I have a few more sums to work out, and I hope to use this strategy in more situations. To subtract the n=0 term (should it be non-zero) from a the result of the standard infinite to get the sum between n=1 and n=infinite makes sense to me, but I've often been wrong about these things before.

14. Nov 15, 2011

### Dick

You are getting pretty close. sum n^2*x^n-sum n*x^n=2x^2/(1-x)^3, right? But I think you've got the sum n*x^n part wrong. Can you go back and fix that?

15. Nov 15, 2011

### Ray Vickson

If you think you will be having trouble at small n, just write out the first few terms explicitly; for example, $S =\sum_{n=0}^\infty x^n = 1 + x + x^2 + \sum_{n=3}^\infty x^n.$ Now it is perfectly clear what the first few terms of $dS/dx \mbox{ and } d^2S/dx^2$ are.

RGV

16. Nov 15, 2011

### Beer-monster

Sorry, I realise now I wrote down my sum wrong on my paper.

Using
$$\sum nx^{n-1} = \frac{1}{(1-x)^{2}}$$

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

$$\frac{2x^{2}}{(1-x)^{3}}+\frac{x^{3}}{(1-x)^{2}}$$

Last edited: Nov 15, 2011
17. Nov 15, 2011

### Dick

Closer and closer. Now you've got sum n*x^n ok. But you don't really mean x^3, do you? Keep checking.

18. Nov 15, 2011

### Beer-monster

No, the x^2s cancel. This is becoming a vivid reminder of how I shouldn't try and rush maths in my head and should sit down with a pencil and shut up :)

19. Nov 15, 2011

### Dick

20. Nov 16, 2011

### Beer-monster

$$\frac{2x^{2}}{(1-x)^{3}}+\frac{x}{(1-x)^{2}}$$

Which can be simplified to:

$$\frac{x(1+x)}{(1-x)^{3}}$$

Which makes sense in the problem I'm doing, and maple agrees :)