# Combining sums problem

1. Oct 3, 2010

### jegues

1. The problem statement, all variables and given/known data

I'm having some confusion about combining sums. Our goal when combining these sums is to have the,

$$(x-c)^{\text{whatever}}$$

term to be the same in both sums.

My confusion is better explained in an example. (see below)

2. Relevant equations

3. The attempt at a solution

Let's say we have the following 2 sums and we want to simplify them into one sum,

$$\sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n+1} + \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n-1}$$

As you can see the,

$$(x-c)^{\text{whatever}}$$

terms are not identical, one is (n+1) and the other is (n-1).

So if we wanted to make the two exponents identical for the first sum we would look as,

$$n \rightarrow n-1$$,

and plug in (n-1) where all the n's used to be in the first sum, and change the starting point of the sum to 1

Now for the second sum, we would look as,

$$n \rightarrow (n+1)$$,

and plug in (n+1) where all the n's used to be in the second sum,

***Here's where I get confused***

But my professor had mentioned to the class that this would not change the starting point of the sum to n= -1, it stays at n=0.

Why is that? Can someone please clarify?

Thanks again!

2. Oct 3, 2010

### jegues

Bump, still trying to figure this one out.

3. Oct 3, 2010

### Inferior89

$$\sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n+1} + \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n-1} =$$

$$\sum_{n=0}^{\infty} (-1)^{n}2^{n}n\left(x^{n+1} + x^{n-1} \right) = \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n+1}\left(x^{-2}+1 \right) = \left(x^{-2}+1 \right) \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n+1}$$

Dunno what else to do..

4. Oct 3, 2010

5. Oct 3, 2010

### Inferior89

Ok, then I don't know.

Btw I don't even see any $$(x-c)^{\text{whatever}}$$ term in any of the sums unless you mean that c = 0. And I made it into one sum with one $$x^{\text{whatever}}$$ where whatever is a function of n.

6. Oct 3, 2010

### Inferior89

Ok. I think the reason it stays at zero is because when you replace n with n+1 and plugin n = -1 for the first term the term will be zero so it doesn't matter if you start your sum at 0 or -1, you will still get the same result.

7. Oct 3, 2010

### jegues

Can anyone else verify this? Is this actually why, or is it just a coincidence?

8. Oct 3, 2010

### Inferior89

Me again! ;)

Well it works with all sums where the first term is zero which they are if you have a sum starting at n=0 of n*f(n,x) and f(n,x) is 'nice' in n=0.