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Combining sums problem

  1. Oct 3, 2010 #1
    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,


    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,

    [tex]\sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n+1} + \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n-1}[/tex]

    As you can see the,


    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,

    [tex]n \rightarrow n-1[/tex],

    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,

    [tex]n \rightarrow (n+1)[/tex],

    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. jcsd
  3. Oct 3, 2010 #2
    Bump, still trying to figure this one out.
  4. Oct 3, 2010 #3
    \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n+1} + \sum_{n=0}^{\infty} (-1)^{n}2^{n}nx^{n-1} =

    [tex]\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} [/tex]

    Dunno what else to do..
  5. Oct 3, 2010 #4

    This doesn't really answer the question I had in the OP.
  6. Oct 3, 2010 #5
    Ok, then I don't know.

    Btw I don't even see any [tex] (x-c)^{\text{whatever}} [/tex] term in any of the sums unless you mean that c = 0. And I made it into one sum with one [tex] x^{\text{whatever}} [/tex] where whatever is a function of n.
  7. Oct 3, 2010 #6
    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.
  8. Oct 3, 2010 #7
    Can anyone else verify this? Is this actually why, or is it just a coincidence?
  9. Oct 3, 2010 #8
    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.
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