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Please critic my proof! (Limit converges to e)

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that 1 + 1 + 1/2! + 1/3! +1/4! + 1/5!... + 1/n!
    Limit of this as n goes to infinity is equal to e.


    2. Relevant equations

    I already showed (1 + 1/n)^n = e

    3. The attempt at a solution

    My proof is as follows:

    e = limit (n to infinity) Sum (k=1 to n) of:

    (n!/(n-k)!) * 1/(k!*n^n)

    = limit (n to infinity) of:

    1 + n/n + n(n-1)/n^2 * 1/2! + n(n-1)(n-2)/n^3 * 1/3! + .........

    Note: each term has the same leading degree and coefficient, meaning each terms tends to 1 * 1/k! as n approaches infinity thus this entire series tends to:

    1 + 1 + 1/2! + 1/3! + 1/4!....

    My main question/concern is the fact that I passed the limit through the summation. It this ok? I know that it works for finite sums:

    ie: Limit (X_n + Y_n) = Limit(X_n) + Limit(Y_n)

    but I'm unsure if this is true for infinite sums. Any help would be appreciated.
     
  2. jcsd
  3. Sep 21, 2009 #2

    Dick

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    You mean sum (k=1 to n) of (n!/(n-k)!) * 1/(k!*n^k), right? You are right to be concerned. No, you can't commute the limit and the sum for infinite series. For example, lim n->infinity sum(k=1 to n) k/n is divergent. sum(k=1 to n) lim n->infinity k/n=0. But your argument does mean it's reasonable to think sum(k=0 to infinity) 1/k! is e. Given you've just proven that lim (n->infinity) (1+1/n)^n is e, that really might be all they are asking for.
     
  4. Sep 21, 2009 #3
    Suppose you're allowed to use the familiar property that [tex]\frac{\mathrm{d} \mathrm{e}^x}{\mathrm{d} x} = \mathrm{e}^x[/tex]. Then first find the Taylor series of the function [tex]\mathrm{e}^x[/tex] and prove that the Lagrange remainder [tex]R_n(x) \to 0[/tex] and thereby showing that [tex]\mathrm{e}^x = \sum_{k = 0}^{\infty} \frac{1}{k!}x^k[/tex] for [tex]\forall x \in \mathbb{R}[/tex]. Set [tex]x = 1[/tex] and that's your result.
     
  5. Sep 21, 2009 #4

    Dick

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    Sure, that works. But I'm not sure this is supposed to use taylor series, is it moo5003?
     
  6. Sep 22, 2009 #5

    lurflurf

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    You are allowed to move that limit though the sum, but first you need to justify it. Once it has been justified you may not need to commute them. A common theme. The usual trick is to break the sum up.
    let n and m be large with m<<n
    then the first m terms of the sum look like those of Σ1/k! and the others are small
     
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