Prove that $\lim_{n\to\infty}c_n=0$ where $c_n=\frac{a_1b_n+a_2b_{n-1}+\cdots+a_nb_1}{n},\lim_{n\to\infty}a_n=0,|b_n|\le B=\text{const}$. Prove that $\lim\limits_{n\to\infty}c_n=0$ where $c_n=\frac{a_1b_n+a_2b_{n-1}+\cdots+a_nb_1}{n},\lim\limits_{n\to\infty}a_n=0,|b_n|\le B=\text{const}$. $\{a_n\},\{b_n\},\{c_n\}$ are sequences.
$$c_n=\lim\limits_{n\to\infty}\left(\frac{a_1b_n}{n}+\frac{a_2b_{n-1}}{n}+\cdots+\frac{a_nb_1}{n}\right)=\lim\limits_{n\to\infty}\frac{a_nb_1}{n}=0$$
Is this correct? 
 A: Since $\text{lim}_{n\rightarrow \infty}a_n=0$ .
Using Cauchy's definition $\forall \epsilon \in \mathbb{R}$  , $\exists N'$ such that for all $n> N'$ ,$a_n<\dfrac{\epsilon}{2 |B|}$
$$|B|\left|\sum_{i=N'}^{n}\dfrac{a_i}{n}\right|< \dfrac{\epsilon} {2}$$
Let $N'=\left\lceil{\dfrac{\text{max}(a_1,a_2,\ldots,a_n)}{\epsilon /2}}\right\rceil$.
Then $\forall n>N''$,
$$\left |\sum_{i=0}^{N''}\dfrac{a_i}{n}\right |<\dfrac{\epsilon}{2|B|} \rightarrow |B|\left|\sum_{i=0}^{N''}\dfrac{a_i}{n}\right |<\dfrac{\epsilon}{2} $$
Let $N = \text{max}(N',N'')$.$\forall n>N,$
$$|c_n| = \left|\dfrac{a_1 b_n+a_2 b_{n-1}+\cdots+a_n b_1}{n}\right |\leq |B|\left|\dfrac{a_1+a_2+\cdots+a_n}{n}\right |=|B|\left(\left|\sum_{i=0}^{N'}\dfrac{a_i}{n}\right |+\left|\sum_{i=N'}^{n}\dfrac{a_i}{n}\right |\right)\\< \dfrac{\epsilon} {2}+\dfrac{\epsilon} {2}=\epsilon$$
Hence $\forall \epsilon \in \mathbb{R}$  , $\exists N$ such that for all $n> N$, $c_n<\epsilon$
$$\therefore \text{lim}_{n\rightarrow \infty}c_n=0$$
A: HINT:
For any $\epsilon>0$ there exists a number $N>0$ such that whenever $n>N$, $-\epsilon<a_n<\epsilon$.
Now, write the sum as 
$$\frac1n \sum_{k=1}^n a_kb_{n-k+1}=\frac1n \sum_{k=1}^Na_kb_{n-k+1}+\frac1n \sum_{k=N+1}^n a_kb_{n-k+1}$$
For the first sum, with fixed $N$, let $n\to \infty$.  For the second sum, use the bounds for $a_n$ and $b_n$.  
A: One the one hand the fact that the number of terms grows means you can't just say that since each term approaches $0$ the sum approaches $0$.  But on the other hand, you're also dividing by $n$ to compensate for that, since the number of terms, although it grows, does not grow faster than $n$.
It is given that for all $n$ we have $|b_n|\le B$.  Therefore
$$
\left|\frac{a_1b_n+a_2b_{n-1}+\cdots+a_nb_1} n\right| \le B \left|\frac {a_1+\cdots+a_n} n \right|.
$$
One question that's been posted here several times is how to prove that
$$
\text{if } \lim_{n\to\infty} a_n = a \text{ then } \lim_{n\to\infty} \frac{a_1+\cdots+a_n} n = a.
$$
I'll see if I can find that and link to it.
