# The avarage of {a(n)}

• May 18th 2010, 08:13 PM
elim
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 $\left\{a_n\right\}$?
• May 18th 2010, 08:40 PM
Drexel28
Quote:

Originally Posted by elim
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 $\left|\frac{a_1+\cdots+a_n}{n}\right|\leqslant\fra c{1}{n}\to 0$
• May 18th 2010, 08:43 PM
mathman88
Quote:

Originally Posted by elim
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?
• May 18th 2010, 08:46 PM
Pinkk
Quote:

Originally Posted by Drexel28
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?
• May 19th 2010, 03:41 PM
elim
Quote:

Originally Posted by mathman88
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 $n>N \Rightarrow$
$\left| \frac{a_1+\cdots+a_n}{n} -A \right| \le \left | \frac{(a_1-A+\cdots + (a_M - A)}{n} + \frac{(a_{M+1}-A)+ \cdots + (a_n-A)}{n}\right |$
$\le \frac{M(\max\{|a_1|,\cdots, |a_M|\}+ |A|)}{n} + \frac{(n-M)}{n} \frac{\epsilon}{2} < \epsilon$