4
$\begingroup$

I'm given that $\displaystyle\lim_{n \to \infty} a_n = a$ and $\displaystyle\lim_{n \to \infty} b_n = b$ and I need to prove that $$\lim_{n\to \infty} \dfrac{a_nb_0+...+a_0b_n}{n} = ab.$$

I'm wondering what are the different kind of proof for this problem.

I know a proof using $\epsilon$ - $N$ definition and I'm interested in more.

So the proof goes like this: Because $a_n$ and $b_n$ converges, we know they are therefore bounded by some constant $M>|a|$. Now, $\forall \epsilon >0$, there exist $N_1$ such that $\forall n>N_1$, $|a_n-a|<\frac{\epsilon}{4M}$ and $|b_n-b|<\frac{\epsilon}{4M}$. Now let $N >max \{N_1, \frac{2M}{\epsilon}[|a_0-a|+...+|a_{N_1}-a|+|b_0-b|+...+|b_{N_1}-b|+|b|]\}$ so then when $n>N$ we have
$ |\frac{a_nb_0+...+a_0b_n}{n} - ab|=|\frac{1}{n}[(a_0b_n-ab)+(a_1b_{n-1}-ab)+...+(a_nb_0-ab)+\frac{ab}{n}|$ =$\frac{1}{n}[(b_n(a_0-a)+a(b_n-b)+b_{n-1}(a_1-a)+a(b_{n-1}-b)+...+b_0(a_n-a)+a(b_0-b)]+\frac{ab}{n}| \leq \frac{M}{n}[|a_0-a|+...+|a_n-a|+|b_0-b|+...+|b_n-b|+|b|] \leq \frac{M}{N}[|a_0-a|+...+|a_{N_1}-a|+|b_0-b|+...+|b_{N_1}-b|+|b|]+\frac{M}{n}[|a_{N_1+1}-a|+...+|a_n-a|+|b_{N_1+1}-b|+...+|b_n-b|]<\frac{\epsilon}{2}+\frac{2M}{n}(n-N_1)\frac{\epsilon}{4M} < \epsilon.$

$\endgroup$
3
  • $\begingroup$ Maybe show the one that you know first in your post. I don't think "$\varepsilon$-$\delta$" is a good classifier. $\endgroup$
    – xbh
    Commented Nov 15, 2018 at 13:28
  • $\begingroup$ @mathnoob : Can you show us the proof using $\epsilon-\delta$? $\endgroup$ Commented Nov 15, 2018 at 13:29
  • $\begingroup$ Stolz theorem may help, but I tried and it didn’t lead to anything useful. $\endgroup$
    – Szeto
    Commented Nov 15, 2018 at 13:56

4 Answers 4

4
$\begingroup$

you can decompose

$$\frac{1}{n+1}\sum_{i=0}^na_ib_{n-i} - ab = \frac{1}{n+1}\sum_{i=0}^n(a_i-a)b_{n-i} + \frac{a}{n+1}\sum_{i=0}^n(b_{i}-b)$$

The first term converges to 0 because $b_i$ are bounded and the second term goes to 0 as well

$\endgroup$
4
  • $\begingroup$ it's $(b - b_i)$ in the second sum right? $\endgroup$
    – mvggz
    Commented Nov 15, 2018 at 13:58
  • $\begingroup$ nice and effective way (+1): I just corrected some typos $\endgroup$
    – G Cab
    Commented Nov 15, 2018 at 18:37
  • $\begingroup$ Not sure your edits are correct GCab ... $\endgroup$
    – Ezy
    Commented Nov 15, 2018 at 19:11
  • $\begingroup$ I edited the edits to make it correct ;) $\endgroup$
    – Ezy
    Commented Nov 15, 2018 at 19:16
1
$\begingroup$

Proof without explicit $\epsilon$-$N$ argument. It suffices to prove that $\frac{1}{n+1}\sum_{i=0}^{n} a_i b_{n-i} \to ab$. Since both $(a_i)$ and $(b_i)$ converge,

  • Both $(a_n)$ and $(b_n)$ are bounded, hence we can pick $M > 0$ so that $|a_n| \leq M$ and $|b_n| \leq M$ for all $n$.

  • If we write $A_n = \sup\{ |a_i - a| : i \geq n\}$ and $B_n = \sup\{ |b_i - b| : i \geq n\}$, then $A_n \to 0$ and $B_n \to 0$.

Then for each fixed $N$ and for each $n \geq N$,

\begin{align*} \left| \frac{1}{n+1}\sum_{i=0}^{n} a_i b_{n-i} - ab \right| &\leq \frac{1}{n+1}\sum_{i=0}^{n} |a_i b_{n-i} - ab| \\ &\leq \frac{M}{n+1}\sum_{i=0}^{n} (|a_i - a| + |b_{n-i} - b|) \\ &\leq \frac{M}{n+1}\sum_{i=0}^{N} (|a_i - a| + |b_{i} - b|) + M(A_N + B_N). \end{align*}

Taking $\limsup$ as $n\to\infty$,

$$ \limsup_{n\to\infty} \left| \frac{1}{n+1}\sum_{i=0}^{n} a_i b_{n-i} - ab \right| \leq M(A_N + B_N). $$

Since the left-hand side is independent of $N$, letting $N \to \infty$ shows that this limsup is zero, hence proves the desired convergence.


A comical twist using probability theory. Let $(\Omega, \mathcal{F}, \mathbb{P}) = \left([0, 1), \mathcal{B}([0,1)), \operatorname{Leb}|_{[0,1)}\right)$ and define $ X_n : \Omega \to \mathbb{R}$ by $ X_n(\omega) = \lfloor (n+1)\omega \rfloor $. Then

  • Each $X_n$ is uniformly distributed over $\{0, \cdots, n\}$,
  • For $\omega \in (0, 1)$, we have $X_n(\omega) \to \infty$ and $n-X_n(\omega) \to \infty$.

Now we note that $ \frac{1}{n+1}\sum_{i=0}^{n} a_i b_{n-i} = \mathbb{E}[a_{X_n}b_{n-X_n}]$. Since $a_{X_n}b_{n-X_n}$ is bounded and converges to $ab$ for $\mathbb{P}$-a.e. $\omega$ (in fact, for all $\omega \in (0, 1)$), the bounded convergence theorem tells that

$$ \lim_{n\to\infty} \frac{1}{n+1}\sum_{i=0}^{n} a_i b_{n-i} = \mathbb{E}\left[ \lim_{n\to\infty} a_{X_n}b_{n-X_n} \right] = \mathbb{E}[ab] = ab. $$

$\endgroup$
0
$\begingroup$

Maybe not so bright idea, but in case if generating functions $f$ and $g$ of sequences $\{a_n\}_{n=0}^\infty$ and $\{b_n\}_{n=0}^\infty$, respectively, exist, one can show that the result by the general Leibniz rule is equal to $$f(1) \cdot g(1) = ab$$

$\endgroup$
0
$\begingroup$

Let us set $c_n=(a*b)(n)=\sum_{k=0}^{n}a_k b_{n-k}$ and $f(x)=\sum_{n\geq 0}a_n x^n$, $g(x)=\sum_{n\geq 0} b_n x^n$.
The convolution implies $f(x)\cdot g(x)=\sum_{n\geq 0}c_n x^n$ and our assumptions are $$ a=\lim_{n\to +\infty}\operatorname*{Res}_{z=0}\frac{f(z)}{z^{n+1}},\qquad b=\lim_{n\to +\infty}\operatorname*{Res}_{z=0}\frac{g(z)}{z^{n+1}}.$$ Let us say that $h(z)=\sum_{n\geq 0}d_n z^n$ is an approximate polynomial iff $\lim_{n\to +\infty}d_n=0$.
Both $f(x)-\frac{a}{1-x}$ and $g(x)-\frac{b}{1-x}$ are approximate polynomials, $f_1(x)$ and $g_1(x)$, and $$ f(x)g(x) = \frac{ab}{(1-x)^2}+\frac{b f_1(x)}{1-x}+\frac{a g_1(x)}{1-x}+f_1(x)g_1(x), $$

$$ (1-x) f(x)g(x) = \frac{ab}{1-x}+\left(\text{approximate polynomial}\right).$$ By considering the coefficient of $x^n$ in both sides, the last identity implies $\lim_{n\to +\infty}(c_{n+1}-c_n)=ab$, hence $\lim_{n\to +\infty}\frac{c_n}{n}=ab$ follows from Cesàro-Stolz.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .