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Thread: The avarage of {a(n)}

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    The avarage of {a(n)}

    We know that if real sequence $\displaystyle \left\{a_n\right\}$ converges, then $\displaystyle \lim_{n \to \infty} (a_{n+1}-a_n) = 0$ and $\displaystyle \left\{ \frac{a_1+\cdots + a_n}{n}\right\}$ converges.

    Suppose $\displaystyle \lim_{n \to \infty} (a_{n+1}-a_n) = 0$ and $\displaystyle \left\{ \frac{a_1+\cdots + a_n}{n}\right\}$ converges, what one can say about $\displaystyle \left\{a_n\right\}$?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by elim View Post
    We know that if real sequence $\displaystyle \left\{a_n\right\}$ converges, then $\displaystyle \lim_{n \to \infty} (a_{n+1}-a_n) = 0$ and $\displaystyle \left\{ \frac{a_1+\cdots + a_n}{n}\right\}$ converges.

    Suppose $\displaystyle \lim_{n \to \infty} (a_{n+1}-a_n) = 0$ and $\displaystyle \left\{ \frac{a_1+\cdots + a_n}{n}\right\}$ converges, what one can say about $\displaystyle \left\{a_n\right\}$?
    Zip.

    Let $\displaystyle a_n=(-1)^n$.

    Then, $\displaystyle \lim_{n\to\infty}(a_{n+1}-a_n)=\lim\text{ }0$ and clearly $\displaystyle \left|\frac{a_1+\cdots+a_n}{n}\right|\leqslant\fra c{1}{n}\to 0$
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    Quote Originally Posted by elim View Post
    We know that if real sequence $\displaystyle \left\{a_n\right\}$ converges, then $\displaystyle \lim_{n \to \infty} (a_{n+1}-a_n) = 0$ and $\displaystyle \left\{ \frac{a_1+\cdots + a_n}{n}\right\}$ converges.

    Suppose $\displaystyle \lim_{n \to \infty} (a_{n+1}-a_n) = 0$ and $\displaystyle \left\{ \frac{a_1+\cdots + a_n}{n}\right\}$ converges, what one can say about $\displaystyle \left\{a_n\right\}$?
    If $\displaystyle \left\{a_n\right\}$ converges, why does $\displaystyle \left\{ \frac{a_1+\cdots + a_n}{n}\right\}$ converge?
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    Senior Member Pinkk's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Zip.

    Let $\displaystyle a_n=(-1)^n$.

    Then, $\displaystyle \lim_{n\to\infty}(a_{n+1}-a_n)=\lim\text{ }0$ and clearly $\displaystyle \left|\frac{a_1+\cdots+a_n}{n}\right|\leqslant\fra c{1}{n}\to 0$
    How does $\displaystyle \lim (a_{n+1} - a_{n}) = 0$ when $\displaystyle a_{n+1} - a_{n} = 2$ for n odd and $\displaystyle a_{n+1}-a_{n} = -2$ for n even?
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  5. #5
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    Quote Originally Posted by mathman88 View Post
    If $\displaystyle \left\{a_n\right\}$ converges, why does $\displaystyle \left\{ \frac{a_1+\cdots + a_n}{n}\right\}$ converge?
    Actually the latter converges to the same limit.
    Given $\displaystyle \epsilon > 0 $, there is an $\displaystyle M > 0$ such that $\displaystyle n > M \Rightarrow |a_n-A| < \epsilon$
    Let $\displaystyle N = \max \left\{ 2M \max\{|a_1|,\cdots, |a_M|\}+ |A|,M\right\} $, then $\displaystyle n>N \Rightarrow$
    $\displaystyle \left| \frac{a_1+\cdots+a_n}{n} -A \right| \le \left | \frac{(a_1-A+\cdots + (a_M - A)}{n} + \frac{(a_{M+1}-A)+ \cdots + (a_n-A)}{n}\right |$
    $\displaystyle \le \frac{M(\max\{|a_1|,\cdots, |a_M|\}+ |A|)}{n} + \frac{(n-M)}{n} \frac{\epsilon}{2} < \epsilon$
    Last edited by elim; May 19th 2010 at 04:14 PM.
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