Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic analysis

  1. Feb 11, 2010 #1
    I am trying to prove a larger problem, that the sequence (2n^2+n)/(n^2) --> 2

    however, i need something small to prove it which is proving the fact that given n > 2,

    2^n - n > n


  2. jcsd
  3. Feb 11, 2010 #2
    Try induction!
  4. Feb 11, 2010 #3
    Exactly what i was thinking, but i need it in a different form to fit my proof.

    Let e > 0 be given.
    Let N = max{2, 1/n}
    Note that, |1/(2^n - n) - 0| = |1/(2^n - n)| (trying to show later in the proof this is less than 1/n)

    Assume n > N
    then n > 2 (Goal: show that 1/(2^n - n) < (1/n) )
    then since |xn - L| < 1/n
  5. Feb 11, 2010 #4
    I need to show 2^n - n > n

    But I need to start the proof with the assumption that n > 2.
  6. Feb 11, 2010 #5
    you can still use induction to prove it? What do you think prevents you from using induction here?
  7. Feb 11, 2010 #6
    he/she probbably needs to prove it using epsilond delta.(formal proof)
    Last edited by a moderator: Feb 11, 2010
  8. Feb 11, 2010 #7
    Yeah that's it. Except i was just given sequence (2n^2+n)/(n^2) --> 2. And in my idea stage, I have the sequence equal to 1/(2^n - n) < 1/n.. and this is only true if n > 2.
  9. Feb 11, 2010 #8
    you seem to be almost done. where is the problem here! it doesn't have to start with n=1 for you to apply induction, as a matter of fact induction holds even if you start at any k positive integer.
  10. Feb 12, 2010 #9


    User Avatar
    Homework Helper

    If your initial sequence involves only [tex] n^2 [/tex], how do you arrive at a a statement that has [tex] 2^n [/tex] involved?

    As I read it, your initial sequence is

    \frac{2n^2 + n}{n^2}

    If this is correct, just simplify the fraction that gives the nth term and you are 99% of the way to proving the sequence converges to 2. If it isn't correct, you have a typo in your original post on the form of the sequence.
    Last edited: Feb 12, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook