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Changing the running parameter

  1. Nov 22, 2008 #1
    hi there!

    I have the following sum:


    and I wanna let it run from n=0 to infty

    it´s somehow obvious that:

    [tex]\displaystyle{\sum_{n=0}^\infty}(-1)^n\frac{(2z)^n}{(2n)!} - 1=\displaystyle{\sum_{n=1}^\infty}(-1)^n\frac{(2z)^n}{(2n)!}[/tex]

    but how do I prove it? I tried to write n-1 instead n but when I calculated it it wasn´t in a way the result from the equation above.

    Does anyone know how I should do this operation?

    thanks in advance
  2. jcsd
  3. Nov 22, 2008 #2


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    The only difference between [itex]\sum_{n=0}^\infty[/itex] and [itex]\sum_{n=1}^\infty[/itex] is the n= 0 term! What is (-1)0 (2z)0/(2(0))! ? Now do you see why you need that "-1" outside the sum?
  4. Nov 22, 2008 #3
    yeah, yeah, I know why it has to be outside, maybe I have to ask my question more precisely. My qestion is, how to change the running parameter, what is the way one does this so that the -1 comes as result of this change?
  5. Nov 22, 2008 #4


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    Well, a summation is just a whole bunch of additions, so it should be obvious to you that

    [tex]\sum_{n=0}^{\infty} a_n=a_0 + \sum_{n=1}^{\infty} a_n \Rightarrow \sum_{n=1}^{\infty} a_n= \sum_{n=0}^{\infty} a_n-a_0[/tex]

    That is really all there is to it. There is no need to change your running variable.
  6. Nov 22, 2008 #5
    :) ok, tnaks once again for the help
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