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Prove x_n=(1+1/n)^n

  1. Sep 19, 2005 #1
    i need to prove that the sequence is increasing, i.e, x_n+1>=x_n
    which translates into:
    [(n+2)/(n+1)]^(n+1)>=[(n+1)/n]^n

    i tried through induction but got stumbled:
    n=k
    [(k+2)/(k+1)]^(k+1)>= [(k+1)/k]^k

    n=k+1
    [(k+3)/(k+2)]^(k+2)>=[(k+2)/(k+1)]^(k+1)
    now i need to prove the last inequality, and got baffled.
    any help is appreciated, perhaps induction isn't the right way?

    p.s
    i know this sequence converges to e, so spare me the trivial details about this sequence.
    thanks in advance.
     
  2. jcsd
  3. Sep 19, 2005 #2
    The derivative is > 0
     
    Last edited: Sep 19, 2005
  4. Sep 19, 2005 #3
    So how do you show that
    [tex]\ln \left(1+\frac{1}{x} \right) > \frac{1}{x+1}, \, x>0[/tex]
     
  5. Sep 19, 2005 #4
  6. Sep 19, 2005 #5
    Try with induction by blocks (blocks of 2^k).
     
  7. Sep 19, 2005 #6

    VietDao29

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    Homework Helper

    Or I think you can also expand the terms out, something like:
    [tex]x_n = \left( 1 + \frac{1}{n} \right) ^ n[/tex]
    So:
    [tex]x_n = 1 + n \times \frac{1}{n} + \frac{n (n - 1)}{2!} \times \frac{1}{n ^ 2} + \frac{n (n - 1) (n - 2)}{3!} \times \frac{1}{n ^ 3} + ... + \frac{n (n - 1) (n - 2) ... (n - (n - 1))}{n!} \times \frac{1}{n ^ n}[/tex]
    [tex]= 1 + 1 + \frac{1}{2!} \times \frac{n}{n} \times \frac{n - 1}{n} + \frac{1}{3!} \times \frac{n}{n} \times \frac{n - 1}{n} \times \frac{n - 2}{n} + ... + \frac{1}{n!} \times \frac{n}{n} \times \frac{n - 1}{n} \times \frac{n - 2}{n} \times ... \times \times \frac{n - (n - 1)}{n}[/tex]
    [tex]= 1 + 1 + \frac{1}{2!} \times \left( 1 - \frac{1}{n} \right) + \frac{1}{3!} \times \left( 1 - \frac{1}{n} \right) \times \left( 1 - \frac{2}{n} \right) + ... + \frac{1}{n!} \times \left(1 - \frac{1}{n} \right) \times \left(1 - \frac{2}{n} \right) \times ... \times \left(1 - \frac{n - 1}{n} \right)[/tex]
    You can do the same and come up with:
    [tex]x_{n + 1} = 1 + 1 + \frac{1}{2!} \times \left( 1 - \frac{1}{n + 1} \right) + \frac{1}{3!} \times \left( 1 - \frac{1}{n + 1} \right) \times \left( 1 - \frac{2}{n + 1} \right) + ... + \frac{1}{(n + 1)!} \times \left(1 - \frac{1}{n + 1} \right) \times[/tex]
    [tex]\times \left(1 - \frac{2}{n + 1} \right) \times ... \times \left(1 - \frac{n}{n + 1} \right)[/tex]
    So can you say that xn + 1 > xn?
    Viet Dao,
     
    Last edited: Sep 19, 2005
  8. Sep 19, 2005 #7
    vietdao, from this expansion the inequality does follow.
    but the question how did you arrive to it?
    have you used newton's binomial?
     
  9. Sep 20, 2005 #8
    Binomial Theorem: (can be proved inductively)

    [tex](a+b)^n=\sum_{i=0}^n \frac{n!}{i! (n-i)!}a^ib^{n-i}[/tex]

    here a=1, b=1/n
     
  10. Sep 20, 2005 #9
    well thank you all, i should have thought of newton's binomial.
     
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