Proof of arithmetic mean of a sequence given the limit of the sequence

Question

Prove that if $\lim_{n\to \infty}a_n=c$ then $\lim_{n\to \infty}A_n=c$ where $A_n$ is the arithmetic mean of the first $n$ of $a_n$. I was only able to prove that is true for a sequence of $1$.

Answer 1

hint

Let $\epsilon>0$ . for large enough $n$, put $b_n=a_n-c $.

$$A_n-c=\frac{b_1+b_2+...+b_N}{n}+\frac {b_{N+1}+...+b_n}{n}$$

where $N $ is such that for $n>N$,

$|b_n|<\epsilon.$

thus

$$|A_n-c|<u_n+\frac{n-N}{n}\epsilon $$ $$<u_n+\epsilon $$

observe that $\lim_{n\to \infty}u_n=0$.

Answer 2

Write $A_n =\frac1{n}\sum_{i=1}^n a_i $ so $A_n-c =\frac1{n}\sum_{i=1}^n (a_i-c) $.

Since $a_i \to c$, for any $\epsilon > 0$ there is a $n(\epsilon)$ such that $i > n(\epsilon) \implies |a_i-c| < \epsilon$.

We now split the sum in $A_n$ into an initial "bad" part and a final "good" part. In the bad part, the $a_i$ are not close to $c$; in the good part they are. The key is that, for any $\epsilon$, there are only a finite number in the bad part, while the good part can grow arbitrarily large, overwhelming the bad part.

To be precise:

$\begin{array}\\ |A_n-c| &=|\frac1{n}\sum_{i=1}^n (a_i-c)|\\ &\le\frac1{n}\sum_{i=1}^n |a_i-c|\\ &\le\frac1{n}(\sum_{i=1}^{n(\epsilon)} |a_i-c|+\sum_{i=n(\epsilon)+1}^n |a_i-c|)\\ &=\frac1{n}\sum_{i=1}^{n(\epsilon)} |a_i-c|+\frac1{n}\sum_{i=n(\epsilon)+1}^n |a_i-c|\\ &\le\frac1{n}n(\epsilon)\max |a_i-c|_{i=1}^{n(\epsilon)}+\frac1{n}\sum_{i=n(\epsilon)+1}^n \epsilon\\ &\le\epsilon+\epsilon \qquad\text{for } n > \dfrac{n(\epsilon)\max |a_i-c|_{i=1}^{n(\epsilon)}}{\epsilon}\\ &\le 2\epsilon \qquad\text{for all large enough } n\\ \end{array} $

Note: The fact that the bound is $2\epsilon$ instead of $\epsilon$ does not matter. Any constant multiple of $\epsilon$ will do.

Answer 3

Let $\epsilon > 0$. Then by the hypothesis there is a natural number $N$ such that for all $n > N$, $|a_n - c| < \epsilon$. Let $K = \sum_{i=1}^{N} a_n$. Then for $n > N$, we have $$\frac{K + (n - N)(c - \epsilon)}{n} < \frac{1}{n}\sum_{i=1}^n a_n < \frac{K + (n - N)(c + \epsilon)}{n},$$ so we have $$\left\lvert c - \frac{1}{n}\sum_{i=1}^n a_n \right\rvert < \left|\frac{cN - K}{n}\right| + \left| \frac{(n-N)\epsilon}{n}\right|$$ by the triangle inequality. Now, the left fraction has constant numerator and thus is less than $\epsilon$ for sufficiently large $n$. The right fraction is always less than $\epsilon$. Thus for sufficiently large $n$ the right hand side of the expression is less than $2\epsilon$, and therefore the arithmetic mean also converges to $c$.