The avarage of {a(n)}

Mar 2010
91
16
usa
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 \(\displaystyle \left\{a_n\right\}\)?
 

Drexel28

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Nov 2009
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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 \(\displaystyle \left|\frac{a_1+\cdots+a_n}{n}\right|\leqslant\frac{1}{n}\to 0\)
 
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Dec 2008
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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?
 
Mar 2009
419
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Uptown Manhattan, NY, USA
Zip.

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

Then, \(\displaystyle \lim_{n\to\infty}(a_{n+1}-a_n)=\lim\text{ }0\) and clearly \(\displaystyle \left|\frac{a_1+\cdots+a_n}{n}\right|\leqslant\frac{1}{n}\to 0\)
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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Mar 2010
91
16
usa
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 \(\displaystyle n>N \Rightarrow\)
\(\displaystyle \left| \frac{a_1+\cdots+a_n}{n} -A \right| \le \left | \frac{(a_1-A+\cdots + (a_M - A)}{n} + \frac{(a_{M+1}-A)+ \cdots + (a_n-A)}{n}\right |\)
\(\displaystyle \le \frac{M(\max\{|a_1|,\cdots, |a_M|\}+ |A|)}{n} + \frac{(n-M)}{n} \frac{\epsilon}{2} < \epsilon\)
 
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