# The avarage of {a(n)}

#### elim

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

MHF Hall of Honor
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 $$\displaystyle \left|\frac{a_1+\cdots+a_n}{n}\right|\leqslant\frac{1}{n}\to 0$$

• mathman88

#### mathman88

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?

#### Pinkk

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?

• Drexel28

#### elim

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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