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Convergence of Series (Harder)

  1. Apr 29, 2012 #1
    Prove that:
    [tex] (1-\frac{1}{n})^n \rightarrow \frac{1}{e} [/tex] as [tex] n \to \infty [/tex]

    you may use the fact that

    [tex] (1+\frac{1}{n})^n \rightarrow e [/tex]

    I have no idea where to even begin, can someone point me in the right direction ?
     
  2. jcsd
  3. Apr 29, 2012 #2

    sharks

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    Try this: Multiply both sides by e.
    [tex](1-\frac{1}{n})^n .e \rightarrow \frac{1}{e} .e[/tex]
    [tex](1-\frac{1}{n})^n .(1+\frac{1}{n})^n \rightarrow 1[/tex]
    If you can simplify the L.H.S. to get 1 on the R.H.S., then you have proved the convergence of the series.
     
  4. Apr 29, 2012 #3
    Really don't know if you can sub in a sequence like that besides I need to prove it converges to [tex] \frac{1}{e}[/tex] not that it simply converges.
     
  5. Apr 29, 2012 #4

    sharks

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    [tex](1-\frac{1}{n})^n .(1+\frac{1}{n})^n \rightarrow 1[/tex]
    [tex]( (1-\frac{1}{n}).(1+\frac{1}{n})) ^n \rightarrow 1[/tex]
    [tex](1-\frac{1}{n}+\frac{1}{n}-\frac{1}{n^2})^n \rightarrow 1[/tex]
    [tex](1-\frac{1}{n^2})^n \rightarrow 1[/tex]
    [tex]\lim_{n\to \infty}(1-\frac{1}{n^2})^n=1[/tex]
     
  6. Apr 29, 2012 #5

    vela

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    Let ##a_n = \left(1 - \frac{1}{n}\right)^n##. Try determining what ##\log a_n## converges to.
     
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