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How would you prove this little inequality?

  1. Mar 2, 2004 #1
    i'm stuck trying to prove this little inequality:

    (1+ 1/n)^n <= e <= (1+1/n)^n+1
    is there a way to prove this without without resorting to power series for e (because we're not allowed to, and we don't know this yet) and also note that n is a natural number, (positive integer).
     
  2. jcsd
  3. Mar 2, 2004 #2
    Here's a hint:

    [tex]\lim_{n \to \infty} \Big( 1 + \frac{1}{n} \Big)^n = e[/tex]

    cookiemonster
     
  4. Mar 2, 2004 #3
    ummmm

    ok i know tha already i just dont know how to prove it give me hint on hwo to prove it
     
  5. Mar 2, 2004 #4
    Just how formally do you want to prove it?

    It's pretty easy to notice that for [itex]n<\infty[/itex], the left side is less than e. When [itex]n = \infty[/itex], it is exactly equal to e.

    The same holds true for the right side, except that it's always greater than e except when [itex]n = \infty[/itex].

    cookiemonster
     
  6. Mar 3, 2004 #5

    matt grime

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    You do it by power series for (1+x)^n valid when |x|<1 (ie x=1/n)
     
  7. Mar 3, 2004 #6
    Why can you not just reason that (1+1/n)^n has to be less than (1+1/n)^n * (1+1/n), as, until n => infinity, 1+1/n will always be a positive value above * a positive value that will always make the right hand larger.
     
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