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.
 A: 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.
A: 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 $ab$.
A: 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 $n \to \infty$.
A: Assume $x_n \to 0$ and $y_n \to b$. Then it is known $\frac{y_1+y_2+\ldots+ y_n}{n} \to b$ and there exists $K>0$ such that  $\frac{y_1+y_2+\ldots+ y_n}{n} \le K$ for every $n$. 
Since $x_n \to 0$ then for a given $\varepsilon>0$ there exists $m>0$ such that $|x_m|<\frac{\varepsilon}{2K}.$
Now we can write
$\frac{x_1y_n+\ldots + x_ny_1}{n}=\frac{x_1y_n+\ldots + x_m y_{n-m+1}}{n}+\frac{x_{m+1}y_{n-m}+\ldots + x_ny_1}{n} \le \frac{x_1y_n+\ldots + x_m y_{n-m+1}}{n} + \frac{\varepsilon}{2K}K$
$\le \frac{x_1y_n+\ldots + x_m y_{n-m+1}}{n} + \frac{\varepsilon}{2} \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$
because the first term is of fixed length $m$ and it can be made small enough.
If $x_n \to a$ then we can write
$\frac{x_1y_n+\ldots + x_ny_1}{n}=\frac{(x_1-a)y_n+\ldots + (x_n-a)y_1}{n}+a\frac{y_1+\ldots + y_n}{n}$
I think that from here everything is clear.
A: Let $\epsilon_0 >0$. $\epsilon=\epsilon_0/M^*$. $M^*$ will be determined later.
Let's verify that $$|\sum_{p=0}^{n} x_p.y_{n-p}-n.ab|<n\epsilon$$
As $\lim_{n}x_n=a$ and $\lim_{n}y_n=b$, we have $n_0$ and $n_0'$ so that $(n>n_0) \implies (|x_n-a|<\epsilon)$ and $(n>n_0') \implies (|y_n-b|<\epsilon)$.
Let $n>n_0+n_0'$.
We have $$\sum_{p=0}^{n} x_p.y_{n-p}-n.ab = (\sum_{p=0}^{n_0-1} x_p.y_{n-p}-n_0.ab) + (\sum_{p=n_0+1}^{n-n_0'} x_p.y_{n-p}-(n-(n_0+n_0')).ab)+(\sum_{p=n-n_0'+1}^{n} x_p.y_{n-p}-n_0'.ab)$$
We have $|x_p.y_{n-p}-ab|=|x_p.(y_{n-p}-b)+b(x_p-a)|\leq|x_p.(y_{n-p}-b)|+|b(x_p-a)|$
So, in the first term : $$|\sum_{p=0}^{n_0-1} x_p.y_{n-p}-n_0.ab| = |\sum_{p=0}^{n_0-1} (x_p.y_{n-p}-ab)|\leq n_0.M.\epsilon+n_0.b.M$$ where $M=max_{0\leq p\leq n_0}(|x_p|,|x_p-a|)\geq\frac{a}{2}$ (max over a finite number of terms, is < $\infty$).
In the third term, we get the same majoration : $$|\sum| \leq n_0'.M'.\epsilon+n_0'.a.M'$$ where $M'=max_{n-n_0' \leq p\leq n}(|y_{n-p}|,|y_{n-p}-b|)$$
And in the middle term : $$|\sum| \leq (n-(n_0+n_0')).((b+\epsilon).\epsilon+a.\epsilon)\leq (n-(n_0+n_0')).((b+a+1).\epsilon$$ 
Hence, for the initial sum : $$|\sum_{p=0}^{n} x_p.y_{n-p}-n.ab|<\epsilon (n_0 M + n_0' M'+(n-(n_0+n_0'))(a+b+1))+n_0 b M + n_0' a M'$$
Now, dividing by n.
$$|\frac{1}{n}\sum_{p=0}^{n} x_p.y_{n-p}-ab|<\epsilon (\frac{C}{n}+(a+b+1)) + \frac{C'}{n}<(a+b+3)\epsilon<\epsilon_0$$ for n large enough, with M*=(a+b+3).QED
