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.
 A: $\displaystyle \sum \frac {a_i b_{n-i} + ab}{n} = \sum \frac {(a_i - a)(b_{n-i} - b)}{n} +a \sum \frac {b_{n-i}}{n} + b \sum \frac {a_i}{n}$
Taking limits as $n \rightarrow \infty$, we have $\sum\frac {b_{n-i}}{n} \to b, \sum\frac {a_i}{n} \to a$, $\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 $\lim_{n\rightarrow \infty} \sum \frac {a_i b_{n-i}}{n} = ab$.
A: 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 $v_n$.
A: 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 $.
A: 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.
A: 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 $n$. 
