# Different methods to calculate the limit of $\frac{a_nb_0+…+a_0b_n}{n}$

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
$$|\frac{a_nb_0+...+a_0b_n}{n} - ab|=|\frac{1}{n}[(a_0b_n-ab)+(a_1b_{n-1}-ab)+...+(a_nb_0-ab)+\frac{ab}{n}|$$ =$$\frac{1}{n}[(b_n(a_0-a)+a(b_n-b)+b_{n-1}(a_1-a)+a(b_{n-1}-b)+...+b_0(a_n-a)+a(b_0-b)]+\frac{ab}{n}| \leq \frac{M}{n}[|a_0-a|+...+|a_n-a|+|b_0-b|+...+|b_n-b|+|b|] \leq \frac{M}{N}[|a_0-a|+...+|a_{N_1}-a|+|b_0-b|+...+|b_{N_1}-b|+|b|]+\frac{M}{n}[|a_{N_1+1}-a|+...+|a_n-a|+|b_{N_1+1}-b|+...+|b_n-b|]<\frac{\epsilon}{2}+\frac{2M}{n}(n-N_1)\frac{\epsilon}{4M} < \epsilon.$$

• Maybe show the one that you know first in your post. I don't think "$\varepsilon$-$\delta$" is a good classifier. – xbh Nov 15 '18 at 13:28
• @mathnoob : Can you show us the proof using $\epsilon-\delta$? – Yadati Kiran Nov 15 '18 at 13:29
• Stolz theorem may help, but I tried and it didn’t lead to anything useful. – Szeto Nov 15 '18 at 13:56

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

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

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

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

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 $$f(1) \cdot g(1) = ab$$

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.