Is it true that $\lim_{n \to \infty} a_n \to L \implies \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} a_k \to L$?

Question

Given a sequence $a_n : \mathbb N \to \mathbb R$ that converges to $L \in \mathbb R$, is it true that $$ \lim_{n \to \infty} a_n \to L \implies \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} a_k \to L $$ Here is my attempt at a proof (and I would appreciate any feedback on whether it is correct or not).

We are given that, for every $\varepsilon_1 > 0$, there exists a $N_1 \in \mathbb N$ such that whenever $n \geq N_1$, $|a_n - L| < \varepsilon_1$. We then want to show that, for every $\varepsilon_2 > 0$, there exists a $N_2 \in \mathbb N$ such that whenever $n \geq N_2$, $\left|\frac{1}{n} \sum_{k=0}^{n-1} a_k - L\right| < \varepsilon_2$.

Fix any $\varepsilon_2 > 0$. Then, choose $\varepsilon_1 = \varepsilon_2$ and, after picking the corresponding $N_1$, choose $N_2 \geq N_1$. Then, $$ \begin{align} \left|\frac{1}{n} \sum_{k=0}^{n-1} a_k - L\right| &\leq \frac{1}{n} \sum_{k=0}^{n-1} \left|a_k - L\right| \\ &< \frac{1}{n} \sum_{k=0}^{n-1} \varepsilon_1 \\ &= \frac{1}{n} \sum_{k=0}^{n-1} \varepsilon_2 \\ &= \varepsilon_2 \cdot \frac{1}{n} \cdot \sum_{k=0}^{n-1} 1 \\ &= \varepsilon_2 \cdot \frac{1}{n} \cdot n \\ &= \varepsilon_2 \end{align} $$

@Bcpicao It is not correct. $|a_k-L|\le \varepsilon_1$ only if $k \ge N_1$, but it is used here for any $k\ge 0$. — Gary, Jan 10 '24 at 11:07
$\vert a_k - L \vert < \epsilon_1$ if $k \ge N_1$, so the bound $$\sum_{k = 0}^{n - 1} \vert a_k - L \vert < \sum_{k = 0}^{n - 1} \epsilon_1$$ is incorrect — Thành Nguyễn, Jan 10 '24 at 11:07
See https://math.stackexchange.com/questions/533626/convergence-of-the-arithmetic-mean — Gary, Jan 10 '24 at 11:11

score 2 · Accepted Answer · answered Jan 10 '24 at 11:13

Your proof is not correct since you attempted to bound all the differences $\vert a_k - L \vert$ by $\varepsilon_1,$ but the starting point was correct. Thus let $N_1$ be as in your post. Then for every $n > N_1$ one has: $$\vert \frac{1}{n + 1} \sum_{0 \leq k \leq n} x_k - L \vert \leq \vert \frac{1}{n} \sum_{0 \leq k < N_1} x_k \vert + \vert \frac{1}{n} \sum_{N_1 \leq k \leq n} x_k - L \vert \leq \\ \frac{N_1}{n} \max_{0 \leq k < N_1} \vert x_k \vert + \frac{1}{n} \sum_{N_1 \leq k \leq n} \vert x_k - L \vert + \frac{N_1 - 1}{n} L \leq \frac{N_1}{n} \max_{0 \leq k < N_1} \vert x_k \vert + \\ \frac{N_1 - 1}{n} L + \frac{n + 1 - N_1}{n} \varepsilon_1.$$ Since the first two terms converge to $0,$ they can be easily made to be less that $\frac{\varepsilon_1}{2}$ for $n \geq N,$ so for all $n \geq \max\{N_1, N\} =: N_2,$ the entire sum from above will be less that $2 \varepsilon_1,$ which yields the desired convergence. I hope this helps. :)

Is it true that $\lim_{n \to \infty} a_n \to L \implies \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} a_k \to L$?

1 Answers1