Results 1 to 2 of 2

Thread: Trying to pin down a divergence proof

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    114

    Trying to pin down a divergence proof

    Hi,

    I would like to show that the sequence $\displaystyle \{1+(-2)^n\}$ diverges. Here is my attempt.

    The sequence converges to some element $\displaystyle L \in \mathbb{R}$ iff for all $\displaystyle \epsilon > 0,\,\, \exists N \in \mathbb{N}$ such that $\displaystyle n > N \implies |a_n - L| < \epsilon$. My aim is to show that there is an epsilon for which no positive integer N satisfies this requirement.

    Choose $\displaystyle \epsilon = 1$. Then there exists an N such that if n > N then $\displaystyle |1+(-2)^n - L| < 1$.

    Now let k be a positive odd integer such that k > N, then

    $\displaystyle |1+(-2)^{k+1} - L| < 1$
    so $\displaystyle |1+2^{k+1} - L| < 1$

    that is $\displaystyle -1 < 1+2^{k+1} - L < 1$

    so $\displaystyle -2 < 2^{k+1} - L < 0,\,\,\implies L > 2^{k+1}$

    and by a similar argument we can show that if we take k to be some even integer such that k > N then

    $\displaystyle L > 2^{k}$

    But we can always find an even AND odd k such that the following inequalities are false for any real number L.

    However there's something in the back of my mind which I'm not happy with. I don't like how I finished the proof, it doesn't seem quite sound to me. I was wondering if someone could point me in the right direction to finish it off or let me know where I've gone wrong and perhaps suggest a better approach.

    Thanks in advance

    Stonehambey
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2010
    Posts
    422
    It looks fine.

    Observe though that $\displaystyle (s_n)_{n=1}^\infty$ diverges iff $\displaystyle (s_n - c)_{n=1}^\infty$ diverges, so you can subtract off the 1 and get the sequence $\displaystyle ((-2)^n)_{n=1}^\infty$. Then observe that a sequence diverges iff it has a divergent subsequence, so you can restrict to even n, in which case you have just $\displaystyle (4^n)_{n=1}^\infty$, whose divergent character is obvious.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Convergence/Divergence proof
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 2nd 2010, 09:06 PM
  2. Divergence theorem proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 6th 2010, 07:26 AM
  3. divergence proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 9th 2009, 04:04 AM
  4. Help with divergence proof
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 8th 2008, 08:08 PM
  5. ratio test divergence proof
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 1st 2007, 10:14 AM

Search Tags


/mathhelpforum @mathhelpforum