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Is this series bounded?

  1. Mar 24, 2010 #1
    I asked by someone.
    But I can't answer to it.
    [tex]\sum^{\infty}_{1}(-1)^n*(1+\frac{1}{n})^n[/tex]
    Is this series bounded?
    I can't do anything about that.
     
    Last edited: Mar 24, 2010
  2. jcsd
  3. Mar 24, 2010 #2

    jbunniii

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    Do you know what

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

    equals?
     
  4. Mar 24, 2010 #3
    I know this series diverge by lim an=e, not zero.
    But I can't figure out it's bounded.
    Would you give me more informations?
     
    Last edited: Mar 24, 2010
  5. Mar 24, 2010 #4
    The sum of any two consecutive terms goes to zero as O(1/n^2). Since the sum of 1/n^2 is convergent, you can prove that your original series is bounded.
     
  6. Mar 24, 2010 #5
    hmm... Do you mean (1 + 1/(n+1))^(n+1) - (1 + 1/n)^n ≤1/n^2?
    I can not understand it. Give me more information, plz.
     
    Last edited: Mar 24, 2010
  7. Mar 24, 2010 #6
    Correct. You need to prove that there's such N and C that, for all n>N,

    (1 + 1/(n+1))^(n+1) - (1 + 1/n)^n ≤ C/n^2

    You can do that if you write

    (1+1/n)^n = exp(n log (1+1/n))

    and then use Taylor expansion of log.
     
    Last edited: Mar 25, 2010
  8. Mar 25, 2010 #7
    I computed the process you have given.
    log(x)=(x-1) - 1/2*(x-1)^2 + 1/3*(x-1)^3 - ...(correct?)
    So My result is
    exp (1 - 1/2(n+1) + 1/3(n+1)^2 - ...)
    - exp (1 - 1/2n + 1/3n^2 - ...)
    let exp(A(n+1)) - exp((An))
    exp does not linear, I can't do further.
    Help me more...
     
  9. Mar 25, 2010 #8
    [tex]exp(1 - \frac{1}{2n} + \frac{1}{3n^2} + ...) \approx e*( - \frac{1}{2n} + \frac{1}{3n^2}) + ( - \frac{1}{2n} + \frac{1}{3n^2})^2/2 + ...) = e*(-\frac{1}{2n}+\frac{1}{3n^2} + \frac{1}{8n^2} + ...) = e*(-\frac{1}{2n}+\frac{11}{24n^2} + ... )[/tex]

    where ... are terms that go down as 1/n^3 or faster.

    [tex]|\frac{1}{2(n+1)} - \frac{1}{2n}| = \frac{1}{2n(n+1)} < \frac{1}{2n^2}[/tex]
     
  10. Mar 25, 2010 #9
    Oh I see!!!
    Thank you for your advice.
    I think this problem is very hard to solve!!!
     
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