Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rudin's explanation of how rapid the series 1/(n!) converges

  1. May 6, 2013 #1
    In Rudin's Principle's of Mathematical Analysis, Rudin days that we can estimate how fast the series [itex]\sum\frac{1}{n!}[/itex] converges by the following:
    Put
    $$
    s_{n}=\sum_{k=0}^{n}\frac{1}{k!}.
    $$
    Then
    $$
    e-s_{n}
    =\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+\cdots<\frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+\cdots\right)=\frac{1}{n!n}
    $$
    so that
    $$
    0<e-s_{n}<\frac{1}{n!n}.
    $$
    The part that bothers me is
    $$
    \frac{1}{(n+1)!}\left(1+\frac{1}{n+1}+\frac{1}{(n+1)^{2}}+\cdots\right)=\frac{1}{n!n}.
    $$
    Using Maple I was able to see that
    $$
    \frac{1}{(n+1)!}\sum_{k=0}^{\infty}\frac{1}{(n+1)^{k}}=\frac{1}{n!n}
    $$
    but what if I did not have access to anything like Maple or Mathematica. How would I be able to figure out that the equality holds?
     
  2. jcsd
  3. May 6, 2013 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    That sum is a geometric series - do you know what to do with those?
     
  4. May 6, 2013 #3
    Oh. HAHAHAHA. Wow. Okay. I see it now. Thanks.
     
  5. May 6, 2013 #4

    mathman

    User Avatar
    Science Advisor
    Gold Member

    The sum is simply a geometric series with r = 1/(n+1), so the sum = 1/(1-r) = (n+1)/n.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Rudin's explanation of how rapid the series 1/(n!) converges
  1. Sequence (n,1/n) (Replies: 6)

Loading...