Aposol's Calculus Vol.1 6.17 #42

  • Thread starter Masaki
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Homework Statement


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].

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.
 

Answers and Replies

  • #2
tiny-tim
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Hi Masaki! :smile:
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.

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)?​
 
  • #3
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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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