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Math Help - disprove convergence

  1. #1
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    disprove convergence

    Hi,

    I have to show that, given a sequence (x_n) of real numbers with the property |x_n|\geq n for infinitely many n, the sequence

    a_n=\frac{1}{n}\sum_{i=1}^n x_n

    does not converge. I tried a lot, but was not very successful.

    Can I somehow show that a_n can't be a Cauchy sequence?

    Any ideas would be appreciated.

    Thank you,
    mili03
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  2. #2
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    Re: disprove convergence

    Assume that a_n converge and try bounding |a_{n+1}-a_n| from below using the fact that \frac{1}{n}-\frac{1}{n(n+1)}=\frac{1}{n+1}.
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  3. #3
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    Re: disprove convergence

    Quote Originally Posted by emakarov View Post
    Assume that a_n converge and try bounding |a_{n+1}-a_n| from below using the fact that \frac{1}{n}-\frac{1}{n(n+1)}=\frac{1}{n+1}.
    thx for the hint. Assuming a_n converges: For each \varepsilon>0 there exists n_0\in N so that for n\geq n_0
    |a_{n+1}-a_n|<\varepsilon.

    So if I find a lower bound, I am finished.

    Now I should probably take a certain n\geq n_0 with x_{n+1}\geq n+1.

    I have |a_{n+1}-a_n|=|\frac{1}{n+1}\sum_{i=1}^{n+1}x_i-\frac{1}{n}\sum_{i=1}^n x_i|=|\frac{1}{n(n+1)}\sum_{i=1}^n x_i-\frac{1}{n+1}x_{n+1}|

    But I honestly speaking don't see how to be continued.

    LG
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  4. #4
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    Re: disprove convergence

    If \lim_{n\to\infty}a_n=A, then there exists an n_0 such that a_n\le 3A/2 for all n\ge n_0.

    \left|\frac{x_{n+1}}{n+1}-\frac{1}{n+1}\frac{1}{n}\sum_{i=1}^n x_i\right|\ge\frac{|x_{n+1}|}{n+1}-\frac{|a_n|}{n+1}\ge1-\frac{3A}{2(n+1)}\xrightarrow[n\to\infty]{}\ 1

    Needs to be double-checked.
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  5. #5
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    Re: disprove convergence

    thank you, that's it! :-)

    Quote Originally Posted by emakarov View Post
    If \lim_{n\to\infty}a_n=A, then there exists an n_0 such that a_n\le 3A/2 for all n\ge n_0.
    Here we require |a_n|\le\frac{3|A|}{2}.
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