Problem. Let $(a_n)_{n=0}^\infty$ and $(b_n)_{n=0}^\infty$ be sequences of real numbers such that $\displaystyle\lim_{n\to\infty}{a_n}=a$ and $\displaystyle\lim_{n\to\infty}{b_n}=b$. Show that $$\lim_{n\to\infty}\frac{1}{n+1}\sum_{k=0}^{n}{a_kb_{n-k}}=ab.$$
My Attempt. It is trivial that $\displaystyle\lim_{n\to\infty}{a_nb_n}=ab$. I was considering to show this in the following fashion: $$\frac{1}{n+1}\sum_{k=0}^{n}{a_kb_{n-k}}-ab=\frac{1}{n+1}\sum_{k=0}^{n}{(a_kb_{n-k}-ab)}=:c_n.$$ Then it suffices to show $c_n\to 0$ as $n\to\infty$. However, this is a bit difficult as we have $a_kb_{n-k}$ instead of $a_kb_k$, which prohibits us from applying the product limit $ab$. That is the place where I get stuck. Hope anyone have good ideas on this.