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Math Help - Trying to pin down a divergence proof

  1. #1
    Member
    Joined
    Nov 2008
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    114

    Trying to pin down a divergence proof

    Hi,

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

    The sequence converges to some element L \in \mathbb{R} iff for all \epsilon > 0,\,\, \exists N \in \mathbb{N} such that 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 \epsilon = 1. Then there exists an N such that if n > N then |1+(-2)^n - L| < 1.

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

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

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

    so -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

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

    Observe though that (s_n)_{n=1}^\infty diverges iff (s_n - c)_{n=1}^\infty diverges, so you can subtract off the 1 and get the sequence ((-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 (4^n)_{n=1}^\infty, whose divergent character is obvious.
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