Results 1 to 2 of 2

Thread: limit of sequence

  1. #1
    Newbie
    Joined
    Jan 2014
    From
    United States
    Posts
    1

    limit of sequence

    Could someone please show me how to prove that this limit is 0 by the squeeze/sandwich theorem in a way as simple as possible. Im not very good with all the technical math descriptions. The sequence is a=(-4)^n/n!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,769
    Thanks
    3027

    Re: limit of sequence

    Since $\displaystyle (-4)^n$ alternates between negative and positive, an obvious "sandwiching" will be between $\displaystyle \frac{4^n}{n!}$ and $\displaystyle -\frac{4^n}{n!}$. One way to prove that $\displaystyle \frac{4^n}{n!}$ goes to 0 is to show that $\displaystyle \sum \frac{4^n}{n!}$ converges since a necessary condition that a series converge is that the sequence of term go to 0. You can do that by using the "ratio test" or even by noting that the series converges to $\displaystyle e^4$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Oct 26th 2010, 10:23 AM
  2. Limit of sequence
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Dec 19th 2009, 03:44 AM
  3. Limit of a sequence
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 25th 2008, 09:07 AM
  4. limit of a sequence
    Posted in the Calculus Forum
    Replies: 5
    Last Post: Feb 4th 2008, 08:08 AM
  5. Limit of Sequence
    Posted in the Calculus Forum
    Replies: 7
    Last Post: Sep 26th 2006, 09:28 AM

Search Tags


/mathhelpforum @mathhelpforum