Results 1 to 3 of 3

Math Help - Analysis Proof

  1. #1
    Junior Member
    Joined
    Sep 2006
    Posts
    32

    Analysis Proof

    Let E be greater than zero. Show that there is a n in N (natural numbers) such that 1/(2^n) is less than E.

    How would you go about proving this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by MKLyon View Post
    Let E be greater than zero. Show that there is a n in N (natural numbers) such that 1/(2^n) is less than E.

    How would you go about proving this?
    Consider the sequence,
    a_n={1/2^n}
    We need to show,
    lim n--> oo a_n=0
    By definition,
    For all e>0 there is an N such that,
    if n>N then, |a_n|=a_n<e

    We note that,
    0<=a_n<=1/n
    And the limits of the sequenes,
    lim n---> oo 0=0
    lim n---> oo 1/n=0
    Thus, by the Squeeze theorem for sequnces we have,
    lim n---> oo a_n=0
    Q.E.D.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,966
    Thanks
    1785
    Awards
    1
    Here is a second way.

    Lemma: If E>0 there is a positive integer n such that (1/n)<E.
    Proof: (1/E)>0 and the positive integers are not bounded above.
    Therefore, the is an integer n such that n>(1/E) or (1/n)<E.

    Now note that for each n, n<2^n. So (1/2^n)<(1/n)<E.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Analysis proof
    Posted in the Differential Geometry Forum
    Replies: 11
    Last Post: April 29th 2010, 09:49 AM
  2. analysis of a proof
    Posted in the Discrete Math Forum
    Replies: 17
    Last Post: November 21st 2009, 07:59 AM
  3. analysis of proof
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 16th 2009, 06:04 AM
  4. Proof analysis
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 9th 2008, 03:20 AM
  5. Analysis Proof
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 18th 2006, 05:24 PM

Search Tags


/mathhelpforum @mathhelpforum