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Aposol's Calculus Vol.1 6.17 #42

  1. Sep 18, 2013 #1
    1. The problem statement, all variables and given/known data
    If [itex]n[/itex] is a positive integer and if [itex]x > 0[/itex], show that
    [itex]\displaystyle\left(1 + \frac{x}{n}\right)^n < e^x[/itex] and that [itex]\displaystyle e^x < \left(1 - \frac{x}{n}\right)^{-n}[/itex] if [itex]\displaystyle x < n[/itex].

    3. The attempt at a solution
    I have proved the first inequality, but I am confused about the second one. Although I know
    [itex]\displaystyle \left(1 - \frac{x}{n}\right)^{-n} = \left(1 + \frac{x}{n-x}\right)^{n}[/itex],
    but I have no idea for the next steps.
     
  2. jcsd
  3. Sep 19, 2013 #2

    tiny-tim

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    Hi Masaki! :smile:
    No, that will only give you that it's < something (from the first part),

    but you need to prove that it's > something. :redface:

    How did you prove the first part (you can probably adapt the proof)?​
     
  4. Sep 19, 2013 #3
    Hello, tiny-tim,
    Actually, there is a problem just before this one, which requests to show
    (a) [itex]\displaystyle e^x > \sum_{k=0}^{n} \frac{x^k}{k!}[/itex]
    and
    (b) [itex]\displaystyle \sum_{k=0}^{2m-1} \frac{(-1)^k x^k}{k!} < e^{-x} < \sum_{k=0}^{2m} \frac{(-1)^k x^k}{k!}.[/itex]

    I could prove the first part by using (a) and binomial expansion, comparing term by term.
    But I don't understand how to use (b) to prove the second part, because both sides of (b) involve positive and negative terms, which makes it difficult to compare them with the second inequality in question....
     
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