1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit of a Sequence

  1. Aug 30, 2006 #1


    User Avatar
    Homework Helper
    Gold Member

    I have to find out where this sequence converges or if it converges a all:

    [tex] a_n = \sqrt{n} - \sqrt{n^2 - 1} [/tex]

    Now, I cant seem to find a good method to solve this. Would my best bet be to use L'hopitals rule to find the limit of the equivalent function or should I try the squeeze theorem. Thats my question. Thanks for the help.
  2. jcsd
  3. Aug 30, 2006 #2
    It looks like it diverges, so show that it is bounded by a divergent sequence.
  4. Aug 30, 2006 #3


    User Avatar
    Homework Helper
    Gold Member

    ok how the this sound as a complete solution?

    An tends seems to tend toward negative infinity and diverge. I proove this by showing that a function which is always greater than it also tends toward negative infinity. I take the greater function to be the squareroot of n.([tex]f(x) = \sqrt{x}[tex]} This function tends toward negative infinity, so the corressponding sequence, and since this sequence is less than the sequence given, the sequence given will also tend toward negative infinity.

    Of course in my solution i'll use more mathmatical notation, but is that the correct reasoning?
  5. Aug 30, 2006 #4
    Yeah that's right. Except I think you mean the negative square root: [tex]f(x)=-\sqrt{x}[/tex]. And make sure you have in there: [tex]f(n) > a_n[/tex] for all [tex]n > a[/tex] (where you find a).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Limit of a Sequence
  1. Sequences and Limits (Replies: 8)

  2. Limit of a Sequence (Replies: 18)

  3. Limits and Sequences (Replies: 8)

  4. Limit of a Sequence (Replies: 1)