Prove that if an infinite sequence converges to $L$ the average of all those numbers also converges to $L$.

Question

Prove that if an infinite sequence converges to $L$, the average of all those numbers also converges to $L$.

My argument would be that if a sequence converges to some number, there are infinitely many numbers that are infinitely close to the limit, so at some point, the previous average is nothing compared to the average of the infinitely many that come before, and because the average of the following numbers of the sequence gets closer to the limit, at the limit, the average should also converge to $L$.

The problem is that I don't know how to formalize it, so I don't have a rigorous proof of this, which I want. Thanks.

Answer 1

Suppose $(a_n)_{n \in \mathbb{N}}$ is a sequence of numbers and that $\lim_{n \rightarrow \infty} a_n = L$, that is, the sequence converges to $L$. We wish to show ...

Hmm... There's some unclarity in the phrase "the average of all those numbers". We can't write "$\frac{1}{\infty}\sum_{n=1}^\infty a_n$" because that's gibberish. When we write "$\sum_{n=1}^\infty a_n$", what we mean is "$\lim_{N \rightarrow \infty} \sum_{n=1}^N a_n$". That last is only a finite sum (for each $N$ in the sequence of $N$s going to infinity), so "$\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N a_n$" is not gibberish and it means "the limit of the average of the leading segment of the numbers in the sequence as the length of the leading segment goes to infinity".

We wish to show $$ L = \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N a_n \text{.} $$

(This proof of Cesaro's lemma is adapted from this MSE question.) We know that for any $\varepsilon > 0$, there is an $M \in \mathbb{N}$ such that for all $n > M$, $|a_n - L| < \varepsilon$. First, we use the lower bound on $a_n$: $a_n >L- \varepsilon$. Then, for $N > M$, \begin{align*} \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N a_n &= \lim_{\substack{N \rightarrow \infty \\ N > M}} \frac{1}{N} \sum_{n=1}^N a_n \\ &\geq \lim_{\substack{N \rightarrow \infty \\ N > M}} \left( \left( \frac{1}{N} \sum_{n=1}^M a_n \right) + \frac{N-M}{N}(L - \varepsilon) \right) \end{align*} This is a limit of a sum of two terms. The first term is a fixed sum of $a_n$s being divided by a larger and larger $N$, so goes to zero. Since $M$ is fixed (during this limit), $\frac{N-M}{N}$ goes to $1$. Therefore, $$ \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N a_n \geq 0 + 1(L - \varepsilon) = L -\varepsilon \text{.} $$

Repeating the argument using the upper bound, for $n > M$, $a_n < L + \varepsilon$, \begin{align*} \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N a_n &= \lim_{\substack{N \rightarrow \infty \\ N > M}} \frac{1}{N} \sum_{n=1}^N a_n \\ &\leq \lim_{\substack{N \rightarrow \infty \\ N > M}} \left( \left( \frac{1}{N} \sum_{n=1}^M a_n \right) + \frac{N-M}{N}(L + \varepsilon) \right) \\ &= 0 + 1(L + \varepsilon) \\ &= L + \varepsilon \text{.} \end{align*}

Therefore, $\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N a_n = L$.

Answer 2

$a_n \to L$. Choose $N$ st. $n>N$ implies $|a_n-L| < \epsilon$.

$\sum_1^n a_k/n = \sum_1^{N} a_k/n + \sum_{N+1}^{n} a_k/n$

First one goes to $0$ as $n\to\infty$, for the second one, \begin{equation} \sum_{N+1}^{n} (L-\epsilon)/n < \sum_{N+1}^n a_k/n < \sum_{N+1}^{n} (L+\epsilon)/n \end{equation} and by noting $(n-N)/n \to 1$, $\sum a_k/n \to L$.

Answer 3

We are given a sequence $a_n, n = 1, 2, \ldots$, with $a_n \to L$. Define the sequences of averages: $b_n = \frac{1}{n}\sum_{i = 1}^{n} a_i$. We want to show that $b_n \to L$.

Fix $\epsilon > 0$. Since $a_n \to L$, we may choose $N$ such that $|a_n - L| < \epsilon/2$ for all $n \geq N$. We have \begin{align*} |b_k - L| &= \frac{1}{k}\left|a_1 + \cdots + a_k - kL\right| \\ &\leq \frac{1}{k}\sum_{i = 1}^k |a_i - L| \\ &= \frac{1}{k}\left(\sum_{i = 1}^{N - 1} |a_i - L| + \sum_{i = N}^k |a_i - L| \right) \\ &< \frac{1}{k}\sum_{i = 1}^{N - 1} |a_i - L| + \frac{k - N + 1}{k}\frac{\epsilon}{2}. \end{align*} Taking the limit as $k \to \infty$, the first term vanishes since the sum is a constant, and the second term tends to $\epsilon/2 < \epsilon$.

Answer 4

Your main idea is correct more formally you can proceed like this :

Let $\epsilon>0$ since $\alpha_n \to \alpha $ there is an $N_1$ such that

$|\alpha_{n} - \alpha| < \frac{\epsilon}{2}$ for every $n \geq N_1$.

Now for every $n > N_1$ you estimate the difference

$\biggl|\dfrac{\alpha_1+...+\alpha_n}{n}-\alpha \biggr|=\biggl|\dfrac{(\alpha_1-\alpha)+...+(\alpha_{N_1}-\alpha)+(\alpha_{N_{1}+1}+\alpha)...+(\alpha_n-\alpha)}{n}\biggr| \leq$

$\biggl|\dfrac{(\alpha_1-\alpha)+...+(\alpha_{N_1}-\alpha)}{n}\biggr|+\biggl|\dfrac{(\alpha_{N_{1}+1}-\alpha)+...+(\alpha_n-\alpha)}{n}\biggr|\leq$

$\biggl|\dfrac{(\alpha_1-\alpha)+...+(\alpha_{N_1}-\alpha)}{n}\biggr| + \dfrac{n-N_1}{n}\dfrac{\epsilon}{2} \leq \ \biggr|\dfrac{(\alpha_1-\alpha)+...(\alpha_{N_1}-\alpha)}{n}\biggr| + \dfrac{\epsilon}{2}.$ $\ $ $(*)$

By using again the fact that $\alpha_n -\alpha \to 0 $ we can find an $M>0$ such that $|\alpha_n-\alpha| \leq M.$

So we can write the $(*)$ equation like this :

$\biggl|\dfrac{\alpha_1+...+\alpha_n}{n}-\alpha\biggr| \leq \dfrac{N_1M}{n} +\dfrac{\epsilon}{2}$. $\ $ $(**)$

Now we pick an $n_0>N_1$ such that $\dfrac{1}{n_0} <\dfrac{\epsilon}{2}\dfrac{1}{N_1M}.$ $(1)$

And now for every $n \geq n_0$ from $(**)$ and $(1)$ we finally have

$\biggl|\dfrac{\alpha_1+...+\alpha_n}{n}-\alpha\biggr| < \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2} = \epsilon.$

Which means that $\dfrac{\alpha_1+...\alpha_n}{n} \to \alpha.$