Results 1 to 6 of 6

Math Help - Brief limit questions.

  1. #1
    Junior Member
    Joined
    Jun 2011
    From
    Colorado, United States
    Posts
    56

    Brief limit questions.

    The question is:
    Prove that the sequence is monotone and bounded. Then find the limit.

    s_{1}=1 and s_{n+1}= \frac{1}{4}( s_{n}+5)for n \in N.

    { s_{n}} is increasing.

    Are we guessing that s_{n} \leq 2?
    Last edited by CountingPenguins; June 29th 2011 at 09:49 PM. Reason: Figured out the formatting.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,964
    Thanks
    1008

    Re: Brief limit questions.

    You need to use tex tags, not math. And your close tag needs /, not \.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2011
    From
    Colorado, United States
    Posts
    56

    Re: Brief limit questions.

    Thanks for the pointer on the formatting. And I think the Lady Gaga equation at the bottom is HILARIOUS!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Brief limit questions.

    Quote Originally Posted by CountingPenguins View Post
    The question is:
    Prove that the sequence is monotone and bounded. Then find the limit.

    s_{1}=1 and s_{n+1}= \frac{1}{4}( s_{n}+5)for n \in N.

    { s_{n}} is increasing.

    Are we guessing that s_{n} \leq 2?
    s_1=1, s_{n+1}=\frac{1}{4}(s_n+5)

    s_n is monotone increasing:

    Induction on n:

    Say holds for n: s_{n+1}>s_n

    Proving for n+1:

    s_{n+2}>s_{n+1}

    s_{n+2}=\frac{1}{4}(s_{n+1}+5)\overset{?}{>}s_{n+1  }=\frac{1}{4}(s_n+5)

    .
    .
    .
    s_{n+1}>s_n



    s_n is bounded:

    Induction on n:

    Say that s_n<2 for all n is true.

    Proving for n+1:

    s_{n+1}=\frac{1}{4}(s_n+5)<\frac{2}{4}+\frac{5}{4}  =\frac{7}{4}<2

    So, s_n is monotonic increasing and bounded, hence \lim_{n\to\infty}s_n exists.

    We denote \lim_{n\to\infty}s_n=l where l\in\mathbb{R}.

    \lim_{n\to\infty}s_{n+1}=\lim_{n\to\infty}\frac{1}  {4}(s_n+5)

    \lim_{n\to\infty}s_{n+1}=\lim_{n\to\infty}s_n=l

    l=\frac{1}{4}(l+5)

    l=\frac{5}{3}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1

    Re: Brief limit questions.

    Quote Originally Posted by CountingPenguins View Post
    The question is:
    Prove that the sequence is monotone and bounded. Then find the limit.

    s_{1}=1 and s_{n+1}= \frac{1}{4}( s_{n}+5)for n \in N.

    { s_{n}} is increasing.

    Are we guessing that s_{n} \leq 2?
    Another solution:



    s_{n+1}=\frac{1}{4}s_n+\frac{5}{4}, s_1=1

    Define new sequence: b_n :

    b_n:=s_n-\frac{5}{3}

    b_n is geometric sequence with q=\frac{1}{4} and b_1=\frac{-2}{3}.

    Hence, b_n=\frac{-2}{3}\frac{1}{4^{n-1}}.

    So, s_n=b_n+\frac{5}{3}=\frac{-2}{3}\frac{1}{4^{n-1}}+\frac{5}{3}

    s_n=\frac{-2}{3}\frac{1}{4^{n-1}}+\frac{5}{3}

    \lim_{n\to\infty}s_n=\lim_{n\to\infty}\frac{-2}{3}\frac{1}{4^{n-1}}+\frac{5}{3}=

    =0+\frac{5}{3}=\frac{5}{3}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: Brief limit questions.

    Quote Originally Posted by CountingPenguins View Post
    The question is:
    Prove that the sequence is monotone and bounded. Then find the limit.

    s_{1}=1 and s_{n+1}= \frac{1}{4}( s_{n}+5)for n \in N.

    { s_{n}} is increasing.

    Are we guessing that s_{n} \leq 2?
    First observe that if this converges it converges to a root of:

    s=\frac{1}{4}(s+5)

    knowing this greatly simplifies your work (in this case it provides a better guess for the upper bound for one thing).

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Limit Questions
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: January 17th 2010, 03:25 PM
  2. Limit Law Questions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 17th 2009, 06:19 AM
  3. limit questions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 17th 2009, 06:40 PM
  4. 2 Limit Questions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 19th 2009, 08:37 PM
  5. some limit questions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 4th 2006, 03:33 PM

Search Tags


/mathhelpforum @mathhelpforum