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Limit of an sum

  1. Dec 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the limit

    [itex]
    lim_{n \to \infty} \sum_{j=1}^n \frac{b^j}{(j+1)!}
    [/itex]


    2. Relevant equations

    Geometric series sum:

    [itex]
    S=\sum_{j=1}^n r^n
    [/itex]

    [itex]
    S-rS=(1-r)S=1-r^{n+1}
    [/itex]

    [itex]
    S=\frac{1-r^{n+1}}{1-r}
    [/itex]

    [itex]
    S \to \frac{1}{1-r} \,\,\, as \,\,\, n \to \infty
    [/itex]

    if [itex] |r|<1 [/itex]

    3. The attempt at a solution

    [itex]
    b\sum_{j=1}^n \frac{b^j}{(j+1)!}-\sum_{j=1}^n \frac{b^j}{(j+1)!}=-\frac{b}{2}+\frac{b^2}{3}+\frac{b^3}{8}+...
    [/itex]

    I tried to use something similar as when deriving the sum of a geometric series, however was unsucessful. I don't know how to integrate a factorial, so I can't use that approach either. Does anyone have any suggestions?
     
  2. jcsd
  3. Dec 21, 2011 #2

    Dick

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    You can forget about the geometric series. That's no help at all. Suppose the problem were [itex]lim_{n \to \infty} \sum_{j=1}^n \frac{b^j}{(j)!}[/itex]. Could you do that one?
     
  4. Dec 21, 2011 #3

    Deveno

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    is it possible that this might represent the taylor series for some function f at b?

    if so, what function do you think it might be?
     
  5. Dec 22, 2011 #4
    hmmm...it might just be a Taylor series. If this was

    [tex] \sum_{n=0}^{\infty} \frac{b^j}{j!} [/tex]

    then this will be the function

    [tex] exp(b) [/tex]

    The sum runs from n=1, so something like

    [tex] exp(b)-1 [/tex]

    still not quite sure what function would result in such a series.
     
  6. Dec 22, 2011 #5

    SammyS

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    That's a start.

    Keep working with your sum like you have been.

    Here's your sum:
    [itex]\displaystyle \sum_{j=1}^n \frac{b^j}{(j+1)!}=\frac{b}{2!}+\frac{b^2}{3!}+ \frac{b^3}{4!}+\frac{b^4}{5!}+\dots[/itex]​

    Here's what you have for eb:
    [itex]\displaystyle \sum_{n=0}^{\infty} \frac{b^j}{j!}=1+\frac{b}{1!}+\frac{b^2}{2!}+ \frac{b^3}{3!}+\frac{b^4}{4!}+\dots[/itex]​

    What if you divide eb by b?
     
  7. Dec 22, 2011 #6
    Yes, that is exactly it. It is

    [tex] \frac{e^b-1-b}{b} [/tex]

    Thank you.
     
    Last edited: Dec 22, 2011
  8. Dec 22, 2011 #7

    SammyS

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    Recheck your result. I think it's not quite right.
     
  9. Dec 22, 2011 #8
    Yes, silly mistake. I fixed it. See above. Thank you.
     
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