# $\lim_{n \to \infty} \frac{1}{n} (\sum_{k=1}^n x_ky_k)=\lim_{n \to \infty}x_n \lim_{n \to \infty}y_n$

if $$\{x_n\}$$ and $$\{y_n\}$$ converge, then $$\lim_{n \to \infty} \frac{1}{n} \bigg(\sum_{k=1}^n x_ky_k\bigg)=\lim_{n \to \infty}x_n \lim_{n \to \infty}y_n$$

Let's say that $$\{x_n\} \to x$$ and $$\{y_n\} \to y$$.
Intuitively, I can see that for a very large $$n$$, the summation will look very similar to $$\frac{nxy}{n}=xy$$. However, I can't seem to find a proper proof for this (how could I translate this intuitive thought into a formal proof?). I tried using the fact that the sequences are bounded but that leaves me with

$$\inf (x_n)\inf(y_n)\le\lim_{n \to \infty} \frac{1}{n} \bigg(\sum_{k=1}^n x_ky_k\bigg)\le\ \sup (x_n)\sup(y_n)$$

I'd love to see some hints. By the way, I'm only using basic aspects about sequences: Cauchy, Weierstrass theorem, Definition of limits, Squeeze theorem and so on. That is, no derivatives or integrals.

• Well, it's the famous theorem about the arithmetic mean of a sequence which tends to the limit of the sequence. In your case you can put $z_n=x_ny_n$ and note that the limit of $z_n$ is the product of the limits of $x_n$ and $y_n$. See here math.stackexchange.com/questions/155839/… Nov 18, 2020 at 1:17

Given $$\epsilon>0$$ choose $$N\in\Bbb N$$ such that $$k>N\implies|x_ky_k-xy|<\epsilon$$. $$\big|\lim_{n\to\infty}\big(\frac{1}{n}\sum_{k=1}^nx_ky_k\big)-\lim_{n\to\infty}x_ny_n\big|=\big|\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n(x_ky_k-xy)\big|\leq\lim_{n\to\infty}\frac{1}{n+N}\sum_{k=1}^{n+N}|x_ky_k-xy|$$ $$\leq\lim_{n\to\infty}\frac{1}{n+N}(n+N)\epsilon=\epsilon$$ $$\therefore\;\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}|x_ky_k-xy|=xy=\lim_{n\to\infty }x_ny_n$$