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How to show that e converges ?

  1. Aug 5, 2010 #1
    Hi All,

    How do we go about showing the euler's number e converges ?
    Recall that

    e =(1+1/n)^n as n ->infinity

    Some place prove this by showing the sequence is bounded above by 3 and is monotonic increasing, thus a limit exist.

    But I forgot how exactly the proof looks like.

  2. jcsd
  3. Aug 5, 2010 #2
    Here's one way.

    Write down (1+1/n)^n using the binomial formula.

    You'll see that it equals to

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

    From here it's not hard to see that it's monotonic increasing, and it's bounded above by 1 + 1 + 1/2! + 1/3! + ... 1/n!.

    In turn, it's not hard to see that the series [itex]\sum 1/n![/itex] converges, because [itex]1/n! < 2^{-n}[/itex] for all n>3.
  4. Aug 5, 2010 #3


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    Use the fact that
    geometric mean<=arithmetic mean
    consider n+1 numbers where 1 is one of the numbers and the other n are 1+1/n
  5. Aug 5, 2010 #4
    That only proves that it's monotonic.
  6. Aug 5, 2010 #5
    Thanks for the neat proof.

    one comment:

    since we know that [itex]\sum 1/n![/itex] converges, we know e converges because
    e = 1 + [itex]\sum 1/n![/itex] where n->infinity
    Last edited: Aug 5, 2010
  7. Aug 5, 2010 #6
    but first you'd have to prove that the original series converges and that its limit is equal to 1 + [itex]\sum_{n=1}^{\infty} 1/n![/itex] .
  8. Aug 5, 2010 #7


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    By Monotone convergence theorem bounded monotonic series converge.
    Last edited: Aug 5, 2010
  9. Aug 5, 2010 #8
    I don't see it.
  10. Aug 6, 2010 #9


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    (1+1/n)^(n+1) is decreasing thus
  11. Aug 6, 2010 #10


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    You repeated use of "e is convergant" is confusing. Numbers do not converge. It makes sense to ask how to show that the limit [itex]\lim_{n\to\infty} (1+ 1/n)^n[/itex] converges or how to show that [itex]\sum_{n=0}^\infty 1/n![/itex] converges.
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