If $(a_k)_{k=1}^{\infty}$ converges to $a\in\mathbb{R}$, then $\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=1}^{n}a_k=a$.

I want to show that, given a sequence of real numbers $(a_k)_{k=1}^{\infty}$ converges to $a\in\mathbb{R}$, then

$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_k=a.$$

My general idea is this:

We know that $\forall \, \varepsilon> 0, \,\exists \,M\in\mathbb{N}$ s.t. $\forall \, m\geq M$, $|a_m-a|<\varepsilon$.

I want to formally say that

$$\begin{aligned} \frac{1}{n}\sum_{k=1}^{n}a_k &\approx \frac{1}{n}\left(\sum_{k=1}^{m-1}a_k+\sum_{k=m}^{n}a\right)\\ &=\frac{1}{n}\left(\sum_{k=1}^{m-1}a_k+(n-m+1)a\right)\\ \end{aligned}$$ Then, as $n$ gets large, since $\sum\limits_{k=1}^{m-1}a_k$ is just some fixed number, $\frac{1}{n}\sum\limits_{k=1}^{m-1}a_k$ will tend to $0$. Also, $\frac{n-m+1}{n}\to1$, so $\frac{n-m+1}{n}\,a\to a$.

This argument seems to be the more intuitive way to explain the result, but the task is to argue it rigorously.

I'm having trouble figuring out how I can formally write the part where I split the sum in the part from $1$ to $m-1$ and $m$ to $n$, I'm not sure how I can express this idea more formally in terms of an $\varepsilon-N$ type argument, or whether there's an alternative way for me to be able to show this formally.

Solution 1:

This is proven quite straitforward: Let $\epsilon > 0$. Scince $a_k \to a$, we can choose $m \in \mathbb N$, such that $|a_n - a| \le \frac \epsilon 2$ for all $n \ge m$. Then we can choose $N \in \mathbb N$, $N \ge m$, such that

$$ \frac 1 N \sum_{j = 1}^m |a_j| \le \frac \epsilon 2.$$

Then, for $n \ge N$:

$$ \left | \frac 1 n \left(\sum_{j = 1}^n a_j \right) - a \right| \le \frac 1 n \sum_{j = 1}^m |a_j| + \frac 1 n \sum_{j = m}^n |a_j -a | \le \frac \epsilon 2 + \frac{n - m + 1}{n} \frac \epsilon 2 \le \epsilon.$$

However, your intuition is good, keep it in mind. The proof basically is just your intuition in formulas. (Your $a_k \approx a$ becomes $|a_k - a| \le \frac \epsilon 2$)

Solution 2:

The Stolz-Cesaro theorem, in the $\cdot/\infty$ case:

Let $(s_n),(b_n)$ be two sequences of real numbers, and let $b_n$ be strictly monotone and divergent. If the following limit exists: $$\tag{1}\lim_{n\to\infty}\frac{s_{n+1}-s_n}{b_{n+1}-b_n}=L\in\Bbb R$$Then: $$\tag{2}\lim_{n\to\infty}\frac{s_n}{b_n}=L$$

If you take $b_n=n$, and $s_n$ as the partial sums $\sum_{k=1}^na_k$ of your original sequence, then the conditions of the Stolz-Cesaro theorem are satisfied. Then consider the limit $(1)$ - $b_{n+1}-b_n=1,\forall n$, so we get:

$$\lim_{n\to\infty}s_{n+1}-s_n=\lim_{n\to\infty}a_{n+1}=a$$

From $(2)$, this is equal to:

$$\lim_{n\to\infty}\frac{s_n}{b_n}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^na_k$$

And we are done:

$$\lim_{n\to\infty}a_{n+1}=a=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^na_k$$