TITLE: Prove that $\lim\frac{1}{n}\sum_{k=1}^n a_kb_{n+1-k}=(\lim a_n)(\lim b_n)$
QUESTION [10 upvotes]: 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.

REPLY [6 votes]: 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 $v_n$.