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Math Help - The avarage of {a(n)}

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    The avarage of {a(n)}

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

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

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

    Let a_n=(-1)^n.

    Then, \lim_{n\to\infty}(a_{n+1}-a_n)=\lim\text{ }0 and clearly \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 \left\{a_n\right\} converges, then \lim_{n \to \infty} (a_{n+1}-a_n) = 0 and \left\{ \frac{a_1+\cdots + a_n}{n}\right\} converges.

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

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