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Thread: 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 $\displaystyle |x_n|\geq n$ for infinitely many n, the sequence

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

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

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

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

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

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

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

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

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

    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 $\displaystyle \lim_{n\to\infty}a_n=A$, then there exists an $\displaystyle n_0$ such that $\displaystyle a_n\le 3A/2$ for all $\displaystyle n\ge n_0$.
    Here we require $\displaystyle |a_n|\le\frac{3|A|}{2}$.
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