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Math Help - Limit Proof

  1. #1
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    Limit Proof

    let all xn>=0 for all natural numbers n. Assume that sigma(n from 1 to infinity) xn<infinity
    Prove that the lim as n tends to infinity of x
    n=0
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by amm345 View Post
    let all xn>=0 for all natural numbers n. Assume that sigma(n from 1 to infinity) xn<infinity
    Prove that the lim as n tends to infinity of x
    n=0
    Assume not; that is, x_n\to L>0. Because x_n converges, \exists N such that n\geq N implies |x_n-L|<\frac{L}{2}\implies \frac{L}{2}<x_n<\frac{3L}{2}

    Let \sum_{n=1}^{N-1} x_n=K (because after all a finite sum is just a number).

    Now we have \sum_{n=1}^{\infty}x_n=K+\sum_{n=N}^{\infty}x_n>K+  \sum_{n=N}^{\infty}\frac{L}{2}

    So...
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  3. #3
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    I'm sorry, I'm still having troubles seeing where to go next?
    Am I looking to end with a contradiction?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Yes. Since L is a constant, you can pull it out of the sum.

    \sum_{n=N}^{\infty}\frac{L}{2}=\frac{L}{2}\sum_{n=  N}^{\infty}1

    which clearly diverges to \infty. This is a contradiction, so L=0.
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  5. #5
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    I just wanted to show another way to do this.

    Let \sum_{n=1}^{\infty}x_n=K
    You know that \sum_{n=1}^N x_n - \sum_{n=1}^{N-1} x_n = x_N


    So taking the limit:

    \lim_{N\to\infty}x_N=\lim_{N\to\infty}\sum_{n=1}^N x_n - \sum_{n=1}^{N-1} x_n

    But because you know that the sum is finite:
    =K-K=0
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  6. #6
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    Yet another way: Let S_n= \sum_{n=1}^N a_n be the nth partial sum. Then the series converges if and only if the sequence of partial sums converges (that's the definition of convergence of a sequence).

    But then, if a_n does not go to 0, S_{n+1}- S_n= a_{n+1} does not go to 0. That means that S_{n}- S_m cannot go to 0 as n and m go to 0 independently and so {S_n} is not a Cauchy sequence.

    Since every convergent sequence is a Cauchy sequence, {S_n} does not converge and thus \sum_{n=1}^\infty a_n does not converge.
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