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.

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

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 $\lim_{n\rightarrow \infty} \sum \frac {a_i b_{n-i}}{n} = ab$.

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 $v_n$.