Cesaro Mean of Sequences - Convergence

Show that if $(x_n)$ is a convergent sequence then the sequence given by the averages $$y_n = \frac{x_1 + x_2 + \cdots + x_n}{n}$$ also converges to the same limit.

Attempt at Proof.

Since $(x_n)$ converges we can say that for all $m$, such that $m \ge N \Rightarrow |x_n - L|\lt \epsilon$.

Base Case. Let n=1 and we have for all m such that $m \ge N_0 \Rightarrow |x_1 - L|\lt \epsilon$ and also for $m \ge N_1 \Rightarrow |x_{n+1} - L|\lt \epsilon$.

Induction Hypothesis. Assume that for an appropriate choice of $N_2$ we have for all m, $m \ge N_2 \Rightarrow |y_{n} - L|\lt \epsilon$.

Choose $max\{N_1,N_2\}$ such that for all $m \ge max\{N_1,N_2\}$ implies $$|\frac{x_1 + x_2 + \cdots + x_n}{n}- L|+ |x_{n+1} - L|\lt 2\epsilon$$

$$= |\frac{x_1 + x_2 + \cdots + x_n+ n x_{n+1}}{n}-2L |\lt 2\epsilon$$

$$= |\frac{x_1 + x_2 + \cdots + x_n+ n x_{n+1}}{n+1}-L |\le |\frac{x_1 + x_2 + \cdots + x_n+ n x_{n+1}}{2n}-L |\lt \epsilon$$ and $$|\frac{x_1 + x_2 + \cdots + x_n+ x_{n+1}}{n+1}-L |\lt|\frac{x_1 + x_2 + \cdots + x_n+ n x_{n+1}}{n+1}-L| \lt \epsilon.$$

Is this approach correct? If not, can you please provide the correct proof. I feel as though I made a mistake in the calculations. Thanks in advance.

Solution 1:

Your approach is wrong. Induction cannaot be used here unless you can get $N$ depending only on $\epsilon $ and not on $n$. Here is a correct proof: $|y_n-L|=|\frac {x_1-L+x_2-L+...+X_n-L} n|\leq \frac {|x_1-L|+|x_2-L|+...+|X_n-L|} n$ Split this into two sums: $\frac {|x_1-L|+|x_2-L|+...+|X_k-L|} n +\frac {|x_{k+1}-L|+|x_{k+2}-L|+...+|X_n-L|} n$ Choose $k$ such that $|x_i-L|<\epsilon $ for $i>k$. Then the second term is less than $\frac {\epsilon +\epsilon +... +\epsilon} n=\frac {n-k} n \epsilon <\epsilon $. The first term tends to $0$ as $n \to \infty $ (because the numerator does not depend on $n$). We are done.

Solution 2:

The intuition for a correct proof of this fact is as follows:

For any $\epsilon>0$, there exists $N$ so that $\lvert x_n-L\rvert<\epsilon$ for all $n\geq N$. Equivalently, we can break the sequence into two parts:

1. Some initial segment $x_1,x_2,\ldots, x_{N-1}$ of terms that can be anything (but there is only a fixed number of them), and

2. A tail $x_{N+1},x_{N+2},\ldots$ of terms that are all close to (read: within $\epsilon$ of) $L$.

If you pick some giant $n$, you get $$ \frac{x_1+x_2+\cdots+x_n}{n}=\frac{x_1+\cdots+x_N}{n}+\frac{x_{N+1}+x_{N+2}+\cdots+x_n}{n}. $$ The first term has a fixed numerator, so it tends to $0$ as $n\to\infty$. (The average makes those few initial terms meaningless in the long run.) The second term can easily be seen to satisfy $$ (L-\epsilon)\frac{n-N}{n}\leq\frac{x_{N+1}+x_{N+2}+\cdots+x_n}{n}\leq(L+\epsilon)\frac{n-N}{n}, $$ and those bounds tend to $L-\epsilon$ and $L+\epsilon$ as $n\to\infty$. (Taking the average of a bunch of terms that are close to $L$ should yield a result close to $L$.)

Can you use these ingredients to complete a proof of the result? The intuition is all there.