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.
Best Answer
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.