Convergence of the sequence given by Cauchy product 
*

*Let $a_n$ and  $b_n$ be two sequences with limit 0.
Does the sequence $c_n = \sum_{k=0}^{n} a_k b_{n-k}$
converge?

*How about if $a_n = t^n$ with $|t| <1$ ?
 A: *

*Not necessarily. Consider $a_n = b_n = \frac{1}{\sqrt[3]{n + 1}}$. Then,
\begin{align*}
c_n &= \sum_{k=0}^n \frac{1}{\sqrt[3]{(k+1)(n - k + 1)}} \\
&= \sum_{k=0}^n \frac{1}{\sqrt[3]{n + 1 + k(n - k)}} \\
&\ge \sum_{k=0}^n \frac{1}{\sqrt[3]{n + 1 + \frac{n^2}{4}}} \\
&= \frac{n + 1}{\sqrt[3]{n + 1 + \frac{n^2}{4}}} \\
&\to \infty,
\end{align*}
hence $c_n \to \infty$. The inequality above comes from maximising the quadratic $k \mapsto k(n - k)$ over $k \in \Bbb{R}$.

*This does work. Suppose that $b_n \to 0$. Fix $\varepsilon > 0$. Then, there exists an $N \in \Bbb{N}$ such that
$$n < N \implies |b_n| < (1 - |t|)\frac{\varepsilon}{2}.$$
We have, for $n > N$,
\begin{align*}
|c_n| &\le \sum_{k=0}^n |b_n| |t|^{n-k} \\
&= \sum_{k=0}^N |b_n| |t|^{n-k} + \sum_{k=N+1}^n |b_n| |t|^{n-k} \\
&\le M \sum_{k=0}^N |t|^{n-k} + (1 - |t|)\frac{\varepsilon}{2} \sum_{k=N+1}^n |t|^{n-k} \\
&\le M \sum_{k=0}^N |t|^{n-k} + (1 - |t|)\frac{\varepsilon}{2} \sum_{k=0}^n |t|^k \\
&\le M \sum_{k=0}^N |t|^{n-k} + (1 - |t|)\frac{\varepsilon}{2} \sum_{k=0}^\infty |t|^k \\
&= \frac{M|t|^{n - N}(1 - |t|^{N + 1})}{1 - |t|} + \frac{\varepsilon}{2}.
\end{align*}
Note that $\frac{M|t|^{n - N}(1 - |t|^{N + 1})}{1 - |t|} \to 0$ as $n \to \infty$, hence there exists some $K$ such that
$$n > K \implies \frac{M|t|^{n - N}(1 - |t|^{N + 1})}{1 - |t|} < \frac{\varepsilon}{2}.$$
Hence,
$$n > \max\{N, K\} \implies |c_n| < \varepsilon,$$
and thus $c_n \to 0$ as $n \to \infty$.
