# Prove that $\lim\frac{1}{n}\sum_{k=1}^n a_kb_{n+1-k}=(\lim a_n)(\lim b_n)$

Let $\displaystyle \lim_{n \to \infty} a_{n} = a$ and $\displaystyle \lim_{n \to \infty} b_{n} = b$. Prove that: $$\lim_{n \to \infty} u_{n} := \lim_{n \to \infty} \frac{a_{1} b_{n} + a_{2} b_{n - 1} + \cdots + a_{n} b_{1}}{n} = ab.$$ I tried to put $x_{n} \leq u_{n} \leq y_{n}$, but I can’t.

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$\displaystyle \sum \frac {a_i b_{n-i} + ab}{n} = \sum \frac {(a_i - a)(b_{n-i} - b)}{n} +a \sum \frac {b_{n-i}}{n} + b \sum \frac {a_i}{n}$

Taking limits as $n \rightarrow \infty$, we have $\sum\frac {b_{n-1}}{n} = b, \sum\frac {a_i}{n} = a$, $\sum \left| \frac {(a_i-a)(b_{n-i}-b)}{n} \right| \leq \sum \frac {(a_i-a)^2 + (b_{n-i}-b)^2} {2n}$, which tends to 0.

Hence, $\lim_{n\rightarrow \infty} \sum \frac {a_i b_{n-i} + ab}{n} = 2ab$, which is equivalent to $\lim_{n\rightarrow \infty} \sum \frac {a_i b_{n-i}}{n} = ab$.

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I like this as it avoids $\delta, \epsilon$. :) – gnometorule Jan 8 '13 at 2:33
@gnometorule $\delta, \epsilon$ was my first instinct actually, but then it got ugly arguing about the middle like the above solutions, so I decided to try modifying the splitting trick. – Calvin Lin Jan 8 '13 at 2:36
This is a very neat argument. +1. You could directly conclude that $$\sum \dfrac{(a_i-a)(b_i-b)}n \to 0$$ since $(a_i-a)(b_i-b) \to 0$, just like you have concluded $\sum \dfrac{a_i}n \to a$ and $\sum \dfrac{b_i}n \to b$. – user17762 Jan 8 '13 at 2:40
@julien I agree with you. I've corrected it now. – Calvin Lin Jan 8 '13 at 2:46
@Marvis Due to the indices being $i$ and $n-i$, I'm not certain / immediately obvious that we can switch the summation signs. We know that $\lim_{n \rightarrow \infty} (a_i - a)(b_{n-i} - b) = 0$, but can't do the sum the other way. – Calvin Lin Jan 8 '13 at 2:49

This solution is definitely far from neat (Calvin's solution is a model of supreme elegance!), but I was more interested in finding explicit $\epsilon$-$\delta$ values.

• Fix $\epsilon > 0$.

• Let $M_{a}$ and $M_{b}$ be upper bounds for the sequences $(|a_{n}|)_{n \in \mathbb{N}}$ and $(|b_{n}|)_{n \in \mathbb{N}}$ respectively.

• Let $N \in \mathbb{N}$ be sufficiently large so that $$\forall i,j \in \mathbb{N}_{> N}: \quad |a - a_{i}| M_{b} < \frac{\epsilon}{6} \quad \text{and} \quad |a||b - b_{j}| < \frac{\epsilon}{6}.$$

• It follows that \begin{align} (*) \quad \forall i,j \in \mathbb{N}_{> N}: \quad |ab - a_{i} b_{j}| &= |(ab - a b_{j}) + (a b_{j} - a_{i} b_{j})| \\ &\leq |ab - a b_{j}| + |a b_{j} - a_{i} b_{j}| \\ &= |a||b - b_{j}| + |a - a_{i}||b_{j}| \\ &\leq |a||b - b_{j}| + |a - a_{i}| M_{b} \\ &< \frac{\epsilon}{6} + \frac{\epsilon}{6} \\ &= \frac{\epsilon}{3}. \end{align}

• Let $n$ be an integer $> 2N$.

• Then \begin{align} \left| ab - \sum_{i=1}^{n} \frac{a_{i} b_{n - i + 1}}{n} \right| = &\left| \frac{nab}{n} - \sum_{i=1}^{n} \frac{a_{i} b_{n - i + 1}}{n} \right| \\ = &\left| \sum_{i=1}^{n} \frac{ab - a_{i} b_{n - i + 1}}{n} \right| \\ \leq &\left| \sum_{i=1}^{N} \frac{ab - a_{i} b_{n - i + 1}}{n} \right| + \\ &\left| \sum_{i=N+1}^{n-N} \frac{ab - a_{i} b_{n - i + 1}}{n} \right| + \\ &\left| \sum_{i=n-N+1}^{n} \frac{ab - a_{i} b_{n - i + 1}}{n} \right|. \end{align}

• Observe that \begin{align} \left| \sum_{i=1}^{N} \frac{ab - a_{i} b_{n - i + 1}}{n} \right| \leq &\sum_{i=1}^{N} \frac{|ab - a_{i} b_{n - i + 1}|}{n} \\ = &\frac{1}{n} \sum_{i=1}^{N} |ab - a_{i} b_{n - i + 1}| \\ \leq &\frac{1}{n} \sum_{i=1}^{N} (|ab| + |a_{i} b_{n - i + 1}|) \\ = &\frac{1}{n} \sum_{i=1}^{N} (|ab| + |a_{i}||b_{n - i + 1}|) \\ \leq &\frac{1}{n} \sum_{i=1}^{N} (|ab| + M_{a} M_{b}) \\ = &\frac{1}{n} \cdot N(|ab| + M_{a} M_{b}). \end{align}

• Hence, if $n > \dfrac{3N(|ab| + M_{a} M_{b})}{\epsilon}$, we get $$\left| \sum_{i=1}^{N} \frac{ab - a_{i} b_{n - i + 1}}{n} \right| < \frac{\epsilon}{3},$$ and by symmetry, we also obtain $$\left| \sum_{i=n-N+1}^{n} \frac{ab - a_{i} b_{n - i + 1}}{n} \right| < \frac{\epsilon}{3}.$$

• Next, notice that \begin{align} \left| \sum_{i=N+1}^{n-N} \frac{ab - a_{i} b_{n - i + 1}}{n} \right| \leq &\sum_{i=N+1}^{n-N} \frac{|ab - a_{i} b_{n - i + 1}|}{n} \\ = &\frac{1}{n} \sum_{i=N+1}^{n-N} |ab - a_{i} b_{n - i + 1}| \\ < &\frac{1}{n} \sum_{i=N+1}^{n-N} \frac{\epsilon}{3} \quad (\text{By $(*)$ above.}) \\ = &\frac{1}{n} \cdot (n - 2N) \cdot \frac{\epsilon}{3} \\ < &\frac{\epsilon}{3}. \end{align}

• Collecting all of these partial results, we arrive at $$\forall n \in \mathbb{N}: \quad n > \max \left( 2N,\dfrac{4N(|ab| + M_{a} M_{b})}{\epsilon} \right) \quad \Longrightarrow \quad \left| ab - \sum_{i=1}^{n} \frac{a_{i} b_{n - i + 1}}{n} \right| < \epsilon.$$

Conclusion: As $\epsilon$ is arbitrary, we have proven that $\displaystyle \lim_{n \to \infty} \sum_{i=1}^{n} \frac{a_{i} b_{n - i + 1}}{n} = ab$.

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This is a generalization of a Cesaro mean which can probably be found somewhere on MSE. Admitting this, we know that $$\lim_{n\rightarrow +\infty}\frac{b_1+\cdots+b_n}{n}=\lim_{n\rightarrow +\infty}\frac{B_n}{n}=b.$$

Now it only remains to show that $\lim_{n\rightarrow +\infty} v_n=0$ where $$v_n=u_n-a\cdot\frac{B_n}{n}=\frac{a_1b_n+\cdots+a_nb_1}{n}-\frac{ab_n+\cdots+ab_1}{n}=\frac{1}{n}\sum_{k=1}^{n}(a_k-a)b_{n-k+1}.$$

Since $b_n$ converges, $|b_n|$ is bounded, say, by $M$. Then we have $$|v_n|\leq \frac{M}{n}\sum_{k=1}^{n}|a_k-a|$$ for all $n$. Another application of the Cesaro mean shows that the RHS converges to $0$, and so does $v_n$.

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The idea is to consider only the middle. Using the definition of convergence, pick some large $m$, such that $a_k, k > m$ is almost $a$ and $b_k, k > m$ is alomst $b$. Take $n$ much larger than $m$. Then you have (1) $n-2m$ terms that are almost $ab$, (2) some constant to represent the remaining $2m$ terms and (3) something small to represent the almost. (1) goes to $ab$ as $n \to \infty$. (2) goes to $0$ as well. (3) goes to $0$ as $m$ goes to infinity. So essentially you have two limits now.

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Hint: subtract $-ab$ from the left side, gathering terms as $t_i := a_i b_{n-i} - ab$ on numerator, and show it goes to $0$. To do this, note (i) there is a $k_0$ such that for $i, n-i > k_0$ the term $t_i$ will be smaller than an $\epsilon$ you choose, (ii) there will be $n - 2k$ such terms smaller than $\epsilon$, and $\frac {n - 2k}{n}$ goes to 1, and (iii) the $k_0$ terms each "on the left and right" of what we control in (i) and (ii) go to $0$ as they can be appropriately bounded in the numerator, and are divided by $n$.

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