Such an example can be constructed from a divergent geometric series.
With $\alpha > \beta > 1$ take
$$a_n = \begin{cases}\,\,\,\,\,\,\,\,\,1 \, , \,\,\,\,\,\,\,\,\,\ n= 0 \\ -\left(\frac{\alpha}{\beta}\right)^n \, , \,\,\, n \geqslant1 \end{cases}$$
and
$$b_n = \begin{cases}\,\,\,\,\,1 \, , \,\,\,\,\,\,\,\,\,\ n= 0 \\ \left(\frac{\alpha}{\beta}\right)^{n-1}\left(\beta^n + \beta^{-(n+1)} \right) \, , \,\,\, n \geqslant1 \end{cases}$$
Since $\alpha/\beta > 1$, the series $\sum a_n$ is a divergent geometric series and $\sum b_n$ diverges by the comparison test.
The general term of the Cauchy product is
$$c_n = a_0b_n + \sum_{k=1}^{n-1} a_{n-k} b_{k} + a_nb_0 \\ = \left(\frac{\alpha}{\beta}\right)^{n-1}\left(\beta^n + \beta^{-(n+1)} \right)- \sum_{k=1}^{n-1}\left(\frac{\alpha}{\beta}\right)^{n-k} \left(\frac{\alpha}{\beta}\right)^{k-1}\left(\beta^k + \beta^{-(k+1)} \right) - \left(\frac{\alpha}{\beta}\right)^n \\ = \left(\frac{\alpha}{\beta}\right)^{n-1} \left(\beta^n + \beta^{-(n+1)} -\sum_{k=1}^{n-1}\left(\beta^k + \beta^{-(k+1)} \right) - \frac{\alpha}{\beta} \right)$$
The finite geometric sums appearing on the RHS are
$$\sum_{k=1}^{n-1}\beta^k = \frac{\beta - \beta^n}{1 - \beta} = \frac{\beta^n - \beta}{\beta -1 } , \\ \sum_{k=1}^{n-1} \beta^{-(k+1)} = \frac{\beta^{-2} - \beta^{-(n+1)}}{1 - \beta^{-1}}= \frac{\beta^{-1} - \beta^{-n}}{\beta - 1}$$
Hence,
$$c_n = \left(\frac{\alpha}{\beta}\right)^{n-1} \left(\beta^n + \beta^{-(n+1)} -\frac{\beta^n - \beta + \beta^{-1} - \beta^{-n}}{\beta -1 } - \frac{\alpha}{\beta} \right)$$
Choosing $\beta = 2$ for convenience we get
$$c_n = \left(\frac{\alpha}{2}\right)^{n-1} \left(2^n + 2^{-(n+1)} -2^n + 2 - 2^{-1} + 2^{-n} - \frac{\alpha}{2} \right) \\ = \left(\frac{\alpha}{2}\right)^{n-1}\left(\frac{3}{2^{n+1}} + \frac{3 - \alpha}{2} \right) \\ = \frac{\alpha^{n-1}}{2^{2n}}\left(3 + (3 - \alpha)2^n \right)$$
Since we require $\alpha > \beta = 2$ we can choose $\alpha = 3$ to get
$$c_n = \left(\frac{3}{4}\right)^n,$$
and the Cauchy product $\sum c_n$ is a convergent geometric series with non-zero terms.
Best Answer
As $(b_n)$ is bounded but divergent, say it as two subsequences converging to two different limits (if not, it would be convergent) $L_1$ and $L_2$. The two corresponding subsequences of $(a_n\cdot b_n)$ will converge to $LL_1$ and $LL_2$ that are different if $L \neq 0$. Thus $(a_n\cdot b_n)$ is divergent.