Results 1 to 4 of 4

Math Help - Question about a "trick" to solve a limit of a sequence ..

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    236

    Question about a "trick" to solve a limit of a sequence ..

    Hi
    i solve a problem about the convergence/divergence of a sequence ..
    and i use the trick which in the attachment ..
    is my solution correct ?
    Attached Thumbnails Attached Thumbnails Question about a "trick" to solve a limit of a sequence ..-sham3a12.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by TWiX View Post
    Hi
    i solve a problem about the convergence/divergence of a sequence ..
    and i use the trick which in the attachment ..
    is my solution correct ?
    I am not entirely sure what the trick you are refering to is. It is true that if \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<  1 that \lim_{n\to\infty}a_n=0. Is that what you are asking?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2009
    Posts
    236
    Yeah ..
    Actually if the series Sigma(an) is convergent then the limit of an = 0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by TWiX View Post
    Yeah ..
    Actually if the series Sigma(an) is convergent then the limit of an = 0
    Yes that is true.

    So for any spectators what is going on is that TWiX is just making the observation that if \sum_{n\in\mathbb{N}}\sigma_n converges it is necessary for \lim_{n\to\infty}\sigma_n. And of course there are a lot of ways to prove that \sum_{n\in\mathbb{N}}\sigma_n converges thaty may be easier than actually computing \lim_{n\to\infty}\sigma_n.

    Example: Compute \lim_{n\to\infty}\frac{(n!)^2}{(2n)!}

    Solution: Note that \lim_{n\to\infty}\left|\frac{\left[(n+1)!\right]^2}{(2(n+1))!}\cdot\frac{(2n)!}{(n!)^2}\right| =\lim_{n\to\infty}\frac{(n+1)^2(n!)^2}{(2n+2)(2n+1  )(2n)!}\cdot\frac{(2n)!}{(n!)^2}=\lim_{n\to\infty}  \frac{(n+1)^2}{(2n+2)(2n+1)}=\frac{1}{4}<1

    Therefore \sum_{n\in\mathbb{N}}\frac{(n!)^2}{(2n)!} converges by the ratio test. Thus by definition \frac{(n!)^2}{(2n)!}\to 0
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to do this algebra "trick"
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: March 1st 2011, 04:53 AM
  2. Replies: 3
    Last Post: October 28th 2010, 10:40 PM
  3. Replies: 2
    Last Post: August 26th 2010, 01:34 PM
  4. Replies: 1
    Last Post: June 4th 2010, 10:26 PM
  5. Sepecial "limit sequence"
    Posted in the Calculus Forum
    Replies: 2
    Last Post: July 18th 2009, 08:00 AM

Search Tags


/mathhelpforum @mathhelpforum