Convergence of a sequence $c_n$

Solution 1:

Counter example:if $a_n,b_n=\frac{(-1)^n}{\sqrt{n}}$,Then $\lim_{n\to \infty}a_n=0$ and $\sum_{i=1}^n\frac{(-1)^i}{\sqrt{i}}$ is convergent (Leibniz test) and we have $c_n=\sum_{k=1}^na_kb_{n-k+1}=(-1)^{n+1}\sum_{k=1}^n\frac{1}{\sqrt{(n-k+1)(k)}}$,now we consider $k(n-k+1)\le(n-k+1)(k+1)=(\frac{n}{2}+1)^2-(\frac{n}{2}-k)^2\le(\frac{n}{2}+1)^2$ and $\mid c_n \mid\ge\frac{2(n+1)}{n+2}$, hence $c_n$ isn't convergent to $0$.

Solution 2:

Suppose moreover that $\displaystyle \sum\limits_{k \geq 1} |b_k|<+ \infty$.

Let $\epsilon>0$ and $N \geq 0$ such that $\displaystyle \sum\limits_{k \geq N } |b_k| <\epsilon$. Also, let $M>0$ be such that $|a_k|<M$ for all $k \geq 0$.

Then $$\left| \sum\limits_{k=1}^n a_{n-k+1}b_k \right| \leq M \sum\limits_{k=N}^n |b_k| + \max\limits_{k \geq n-N+2} |a_k| \cdot \sum\limits_{k=1}^{N-1} |b_k|$$

For $n$ large enough, you get $$\left| \sum\limits_{k=1}^n a_{n-k+1}b_k \right| \leq (M+1)\epsilon$$

Therefore, it is true that $\lim\limits_{n\to + \infty} c_n=0$ when $\displaystyle \sum\limits_{n \geq 1} b_n$ is absolutely convergent.