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Limit of sequence

  1. Nov 30, 2004 #1
    Hi all,

    I can't move with this limit:

    [tex]
    \lim_{n \rightarrow \infty} n \left( \left( 1 + \frac{1}{n} \right)^{n} - e \right)
    [/tex]

    Could someone help me please? Or some hint...

    But no l'Hospital please.

    Thank you
     
  2. jcsd
  3. Nov 30, 2004 #2
    Find out what

    [tex]
    \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^{n}
    [/tex]

    is. It should all be clear from then on.
     
  4. Nov 30, 2004 #3
    Well I know it is equal to e, but then

    [tex]
    \lim_{n \rightarrow \infty} n \left( e - e \right) = \lim_{n \rightarrow \infty} n \left ( 0 \right) = 0.\infty
    [/tex]

    Which is undefined...
     
  5. Nov 30, 2004 #4

    shmoe

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    Are you opposed to using the series expansions of log(1+x) and e^x? You can re-write the (1+1/n)^n term as e^(n*log(1+1/n)). Use the Taylor series for log(1+x), at least 2 terms +remainder. Then you can factor out an e and use the series for e^x, (2 terms+remainder here as well), and that will do it.

    If you are opposed to using series but have learned them, it's probably possible to modify the above into a squeeze limit type of proof. If you haven't learned Taylor series yet, I'll have to think of a more elementary method, though it will probably be a thinly disguised version of the above.
     
  6. Nov 30, 2004 #5
    Thank you shmoe, maybe this limit is really supposed to be solved using Taylor series, because it is from sample calculus test we're gonna take, but as I see in sylabus, we'll learn Taylor series before the test. It confused me, because the sequences are already behind us and I thought I should already be able to solve any limit of sequence....
     
  7. Nov 30, 2004 #6
    you cannot sub in the value for e in between the limit like that
     
  8. Nov 30, 2004 #7
    i wonder if this will work though

    split the limit so you have the limits of n(1-1/n)^n and n e subtracted from each other.

    now the limit of n (1-1/n)^n is infinity while the limit of n E is also infinity

    however the first one converges to e more slowly that E already does

    so n E >n(1-1/n)^n and the limit is negative infinity
     
  9. Nov 30, 2004 #8
    The limit should be -e/2...
     
  10. Nov 30, 2004 #9
    check what you entered into maple/mathematica/matlab... it isn't -e/2, it is -infinity
     
  11. Nov 30, 2004 #10

    shmoe

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    No, this won't work. You're essentially saying infinity-infinity=-infinity because the first infinity is getting there slower. This is bunk. An indeterminate infinity-infinity limit form can potentially equal anything we like(-e/2 is correct in this case, done by hand with the method I suggested).

    twoflower-series is likely what's intended then. I can't see another way that isn't unnecessarily complicated.
     
  12. Nov 30, 2004 #11

    NateTG

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    If you can show that
    [tex]-\frac{e}{2n} - \epsilon_1(n) \leq \left( 1 + \frac{1}{n} \right)^{n} - e \leq -\frac{e}{2n} + \epsilon_2(n)[/tex]

    Where [itex]\epsilon(n)[/itex] is some expression that goes to zero faster than
    [itex]\frac{1}{n}[/itex] then you're set, so perhaps you should look at the convergence of [itex]\left( 1 + \frac{1}{n} \right)^{n} [/itex]
     
  13. Nov 30, 2004 #12
    I tried it now. Maple gives -e/2
     
  14. Nov 30, 2004 #13

    shmoe

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    Nate, do you have a simple method in mind for those inequalities?

    The obvious approach to me involves some bounds on log(1+x) and e^x that I would prove using series.
     
  15. Nov 30, 2004 #14
    Ok, so you say guys that the most natural way to find this limit is to use Taylor series. If it's right, I will skip this one. Or does anyone have simplier approach to show?
     
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