1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Gold Member

    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


    User Avatar
    Gold Member

    [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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Let ##a_n = \left(1 - \frac{1}{n}\right)^n##. Try determining what ##\log a_n## converges to.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Convergence of Series (Harder)
  1. Series Convergence (Replies: 3)

  2. Convergence of series (Replies: 2)

  3. Series convergence (Replies: 26)

  4. Convergence of a series (Replies: 24)

  5. Series Convergence (Replies: 15)