Find the limit $\lim_{n\to\infty} \frac{x_1 y_n + x_2 y_{n-1} + \cdots + x_n y_1}{n}$

When $\lim_{n\to\infty} x_n = a$, and $\lim_{n\to\infty} y_n = b$, find the limit, $$\lim_{n\to\infty} \frac{x_1 y_n + x_2 y_{n-1} + \cdots + x_n y_1}{n}.$$ Thank you for your help in advance.

Solution 1:

We can use the standard result:

If $x_{n} \to x$ as $n \to \infty$ then $$\lim_{n \to \infty}\frac{x_{1} + x_{2} + \cdots + x_{n}}{n} = x$$

This is pretty standard and its proof is available on MSE. Now for the current question let $x_{n} = a + e_{n}$ and then $e_{n} \to 0$ as $n \to \infty$. We have $$\begin{aligned}\frac{x_{1}y_{n} + x_{2}y_{n - 1} + \cdots + x_{n}y_{1}}{n} &= a\cdot\frac{y_{1} + y_{2} + \cdots + y_{n}}{n}\\ &+ \frac{e_{1}y_{n} + e_{2}y_{n - 1} + \cdots + e_{n}y_{1}}{n}\end{aligned}$$ Now in the above the first term tends to $a\cdot b = ab$. For the second term we need to note that the sequence $y_{n}$ is bounded by some $K > 0$ and therefore in absolute value the second term is no greater than $K\cdot\dfrac{|e_{1}| + |e_{2}| + \cdots + |e_{n}|}{n}$. Since $e_{n} \to 0$ therefore this term also tends to $0$. So the desired limit is $ab$.

Solution 2:

By the Cesàro mean theorem, if $(x_n)_{n\in\mathbb{N}^*}\to a$ then $\left(\bar{x}_n=\frac{1}{n}\sum_{j=1}^{n}x_j\right)_{n\in\mathbb{N}^*}\to a$.

So, for any $\epsilon>0$ there exists $N\in\mathbb{N}$ such that all the quantities: $$|x_m-a|,\quad|y_m-b|,\quad|\bar{x}_m-a|,\quad|\bar{y}_m-b|$$ are less than $\epsilon$ for any $m\geq N$. If we set: $$ c_n = \frac{1}{n}\sum_{i=1}^{n}x_i y_{n+1-i}, $$ for any $n\geq N$ we have that: $$ c_{2n} = \frac{1}{2n}\sum_{j=1}^{n} x_j y_{2n+1-j}+\frac{1}{2n}\sum_{j=1}^{n} y_j x_{2n+1-j} $$ differs from $\frac{1}{2}b \bar{x}_n+\frac{1}{2}a \bar{y}_n$ no more than $\frac{\epsilon}{2}(|\bar{x}_n|+|\bar{y}_n|)$, so: $$ \left(c_{2n}\right)_{n\in\mathbb{N}^*}\to ab. \tag{1}$$ In a similar fashion, for any $n\geq N$ $$ c_{2n+1} = \frac{2n}{2n+1}\left(\frac{1}{2n}\sum_{j=1}^{n} x_j y_{2n+2-j}+\frac{1}{2n}\sum_{j=1}^{n} y_j x_{2n+2-j}\right)+\frac{x_{n+1}y_{n+1}}{2n+1} $$ cannot differ from $\frac{2n}{2n+1}\left(\frac{1}{2}b \bar{x}_n+\frac{1}{2}a \bar{y}_n\right)$ more than $\left(\frac{\epsilon}{2}+\frac{\varepsilon^2}{n}\right)\cdot(|\bar{x}_n|+|\bar{y}_n|)$, so: $$ \left(c_{2n+1}\right)_{n\in\mathbb{N}}\to ab. \tag{2}$$ Now $(1)$ and $(2)$ simply give: $$ \left(c_n\right)_{n\in\mathbb{N}^*}\to ab \tag{3}$$ as expected.

Solution 3:

In the special case that $x_n,y_n\ge0$, we can prove the statement as follows: by the Arithmetic Mean-Geometric Mean (first inequality) and the Cauchy-Schwarz ( second inequality) we can bound the given term as follows:

$$\sqrt[n]{x_1\ldots x_ny_1\ldots y_n}\le \dfrac{x_1y_1+\ldots+x_ny_n}{n} \le \dfrac{\sqrt{x_1^2+\ldots+x_n^2}\sqrt{y_1^2+\ldots+y_n^2}}{n}$$

Letting $n \to \infty$ we have that : $$\sqrt[n]{a^nb^n}\le \lim_{n \to \infty} \dfrac{x_1y_1+\ldots+x_ny_n}{n}\le\sqrt{a^2}\sqrt{b^2}\tag{1}$$ or equivalently $$ab\le \lim_{n \to \infty} \dfrac{x_1y_1+\ldots+x_ny_n}{n}\le ab$$

Using (1), if $x_n \to a$, as $n \to \infty,$ then also we should have

$$\prod_{k=1}^{n} x_k^{1/n}\to a \quad \text{as} \quad n\to \infty, $$

since $$\prod_{k=1}^{n}x_k^{1/n}=\exp\left(\frac{1}{n}\sum_{k=1}^{n}\ln x_k\right)$$ and $\exp, \ln$ are continuous, thus enabling taking the limit taken inside. We also used the well known fact that if $x_n \to a$ as $n \to \infty$ then $x_n^2 \to a^2$ and also$$\frac{x^2_1+\ldots+x^2_n}{n} \to a^2$$ as $n \to \infty$.