Results 1 to 5 of 5

Math Help - 2 Sequence Problems

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    48

    2 Sequence Problems

    Find the value of the limit of the sequence defined by:
    a(sub 1)=1
    a (sub n+1)=13-(1/a(sub n))

    I have no idea how to do that.


    Find the value of the limit of the sequence defined by:
    a(sub 1) = Square root of 10
    a(sub n+1) = Square root of (10+a(sub n))

    This one either. These are the only two out of 30 that I couldn't do. He said they'd be on the test so I need to learn... bummer.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by thegame189 View Post
    Find the value of the limit of the sequence defined by:
    a(sub 1)=1
    a (sub n+1)=13-(1/a(sub n))

    I have no idea how to do that.


    Find the value of the limit of the sequence defined by:
    a(sub 1) = Square root of 10
    a(sub n+1) = Square root of (10+a(sub n))

    This one either. These are the only two out of 30 that I couldn't do. He said they'd be on the test so I need to learn... bummer.
    Try this...

    For the last one...

    suppose \lim_{n \to \infty}a_n=L then \lim_{n \to \infty}a_{n+1}=L

    \lim_{n \to \infty}[a_{n+1}=\sqrt{10+a_n}] \iff L=\sqrt{10+L}

    Now we can solve this equation for L

    L=\sqrt{10+L} \iff L^2=10+L \iff L^2-L-10=0

    L=\frac{1}{2}(1+\sqrt{41})
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2008
    Posts
    48
    Ah, I had a sign wrong that I didn't catch. Thanks a bunch!

    Any idea how to do the first one?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member javax's Avatar
    Joined
    Jan 2008
    From
    Milky Way
    Posts
    139
    Quote Originally Posted by TheEmptySet View Post
    Try this...

    For the last one...

    suppose \lim_{n \to \infty}a_n=L then \lim_{n \to \infty}a_{n+1}=L

    \lim_{n \to \infty}[a_{n+1}=\sqrt{10+a_n}] \iff L=\sqrt{10+L}

    Now we can solve this equation for L

    L=\sqrt{10+L} \iff L^2=10+L \iff L^2-L-10=0

    L=\frac{1}{2}(1+\sqrt{41})
    heeeem...I'm also having problem with these sequences! We don't know yet that the limit exists! The first steps are to prove that the sequence is bounded(a bit messy for me) and monotone! Then if it is, comes your step!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    The first sequence

    I'll post the proof that \lim{a_n} exists

    First we'll show that: a_{n+1}>a_n if n\geq{1}

    By induction:

    a_2=12>a_1=1 is true

    Assume a_{n}>a_{n-1} holds then we'll prove that: a_{n+1}>a_n

    a_n>a_{n-1} then -a_{n}<-a_{n-1}
    so: -\frac{1}{a_{n-1}}<-\frac{1}{a_{n}} and then: 13-\frac{1}{a_{n-1}}=a_n<13-\frac{1}{a_{n}}=a_{n+1} which completes the first part

    a_{n+1}>a_n>...>a_1=1>0 so it is a positive sequence and therefore it is bounded by 13 since a_{n+1}=13-\frac{1}{a_n}<13

    This shows that the sequence is incresing and a_n<13 for all n\geq{1}, thus it converges to say L

    LIMIT:
    Since a_{n+1}=13-\frac{1}{a_n} taking the limit on both sides: \lim{a_{n+1}}=13-\frac{1}{\lim{a_n}} then L=13-\frac{1}{L} now find the roots of this equation and remember that 13\geq{L}>12=a_2 to see which one is the limit.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. sequence problems
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 13th 2011, 07:46 PM
  2. A pair of sequence problems
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 7th 2009, 01:38 PM
  3. Sequence Problems
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 5th 2009, 12:04 PM
  4. 3 sequence problems
    Posted in the Calculus Forum
    Replies: 0
    Last Post: September 14th 2009, 09:43 AM
  5. More sequence problems :(
    Posted in the Calculus Forum
    Replies: 9
    Last Post: May 17th 2008, 12:34 PM

Search Tags


/mathhelpforum @mathhelpforum