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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.

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    $\begingroup$ You should guess that the limit is $ab$. $\endgroup$ – user99914 Mar 25 '14 at 8:53
  • $\begingroup$ It's obviously $ab$, but I wonder if it can be proven nicely :/ $\endgroup$ – user2345215 Mar 25 '14 at 8:54
  • $\begingroup$ It can be proved, but probably the one that I have in mind is not a nice one ^^ $\endgroup$ – user99914 Mar 25 '14 at 8:57
  • $\begingroup$ You should be able to use the idea of this question which deals with the case the $y_i$ are $1$. There are several other MSE questions that show that if the limit exists, the limit of the means exists. $\endgroup$ – André Nicolas Mar 25 '14 at 9:01
  • $\begingroup$ Try by determining $N$ so that $|a-x_{n>N}|<\epsilon$ AND the same for y, and then write as sum(first N)+sum(last N)+sum(middle where both conditions apply). The middle part should dominate in the limit. $\endgroup$ – orion Mar 25 '14 at 9:16
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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 $ab$.

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  • $\begingroup$ Very nice! But what if $x_n$ cannot be expressed in the additively separable form? $\endgroup$ – wshito Mar 26 '14 at 9:50
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    $\begingroup$ @life247-1: Whatever $x_{n}$ is, we define a sequence $e_{n}$ as $e_{n} = x_{n} - a$. $\endgroup$ – Paramanand Singh Mar 26 '14 at 11:51
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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.

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  • $\begingroup$ Usually, the terms less than $N$ can be ignored since they are bounded and only has finite number of terms. However, since the both ends (first $N$ and the last $N$ terms) are indexed reversely from 1 to $N$, it is the middle terms that stretches as $n\to \infty$. So, I think we cannot simply conclude that the sum of infinitely many middle terms is bounded, can we? $\endgroup$ – wshito Mar 25 '14 at 17:11
  • $\begingroup$ Infinitely many middle terms? There is at most one middle term, the four initial differences are bounded by $\epsilon$ for any $m\geq N$. $\endgroup$ – Jack D'Aurizio Mar 25 '14 at 19:27
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    $\begingroup$ What I use is $$\left|\left(\sum_{j=1}^{n}x_j y_{2n+2-j}\right)-b\sum_{j=1}^{n}x_j\right|\leq \epsilon n |\bar{x}_n|,$$ for both ends. $\endgroup$ – Jack D'Aurizio Mar 25 '14 at 19:31
  • $\begingroup$ I am talking about the terms for any $n < N$. Aren't there infinitely many terms as $n\to\infty$? How do you prove them to be bounded? $\endgroup$ – wshito Mar 26 '14 at 1:21
  • $\begingroup$ Every Cauchy sequence in $\mathbb{R}$ is bounded. $\endgroup$ – Jack D'Aurizio Mar 26 '14 at 8:12
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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 $n \to \infty$.

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    $\begingroup$ your proof depends on the premise that $x_n,y_n \geq 0$, which is not given.... $\endgroup$ – mengdie1982 Sep 23 '18 at 19:34
  • $\begingroup$ @mengdie1982 Thank you very much for noting this. $\endgroup$ – Jimmy R. Sep 24 '18 at 1:41
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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.

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  • $\begingroup$ But $x_n$ does not tend to $0$. $\endgroup$ – user10444 Mar 25 '14 at 10:07
  • $\begingroup$ Yes it does not. $\endgroup$ – kmitov Mar 25 '14 at 10:08
  • $\begingroup$ This can be used to proof the general case. $\endgroup$ – kmitov Mar 25 '14 at 10:09
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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

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  • $\begingroup$ Your proof assumes $n_0 < n_0'$ and splits the sequence as $$x_1 y_n + \cdots x_{n_0}y_{n-n_0+1} + \cdots + x_{n-n_0' + 1}y_{n_0'} + \cdots + x_n y_1.$$ How about the case for $n_0' < n_0$ as $$x_1 y_n + \cdots + x_{n_0'} y_{n_0'} + \cdots + x_{n_0} y_{n_0} + \cdots + x_n y_1 ?$$ $\endgroup$ – wshito Mar 26 '14 at 1:12

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