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Series sum questio

  1. Jun 14, 2010 #1
    30lch2b.jpg
    they take members from the marked series
    how they make the power to be from n-1 to n+1
    ??

    and why if we take the member of n=0 n=1
    the index starts from n=1 and not n=2

    ??
     
  2. jcsd
  3. Jun 14, 2010 #2
    If I got right your question.

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

    Remember: 0! = 1
    (כתב יפה:smile:)
     
    Last edited: Jun 14, 2010
  4. Jun 17, 2010 #3

    LCKurtz

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    You are apparently expecting the sum you underlined to be:

    [tex]\sum_{n=2}^\infty \frac 1 {n!}\left(\frac 1 u\right)^{n-1}[/tex]

    because the n = 0 and n = 1 terms were written separately. But notice what happens in this sum if you change the variables:

    k = n -1 or n = k + 1. You get

    [tex]\sum_{k=1}^\infty \frac 1 {(k+1)!}\left(\frac 1 u\right)^{k}[/tex]

    which, if you call the index n instead of k, is what is written.
     
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