As Davide Giraudo has said in the comments, we can find a counter example by using $a_k = b_k = (-1)^k/k$ for $k\geq 1$ and $a_0=b_0=0.$ In that case we compute $$c_n = \sum_{k=0}^n a_{n-k} b_k = \sum_{k=1}^{n-1} \frac{(-1)^{n-k}}{n-k} \frac{(-1)^k}{k} = (-1)^n \sum_{k=1}^{n-1} \frac{1}{k(n-k)}$$
$$ = \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{n-k+k}{k(n-k)} = \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \left( \frac{1}{k} + \frac{1}{n-k}\right) = 2\frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{1}{k}.$$
We show $\sum c_n$ converges by applying the Leibniz criterion. $c_n \to 0$ is clear, so we need only verify that $d_n = \frac{1}{n} \sum_{k=1}^{n-1}\frac{1}{k} $ is monotonically decreasing for sufficiently large $n.$
We compute $$d_{n+1}- d_n = \frac{1}{n+1} \sum_{k=1}^n \frac{1}{k} - \frac{1}{n} \sum_{k=1}^{n-1} \frac{1}{k}= \frac{1}{n(n+1)} - \frac{1}{n(n+1)}\sum_{k=1}^{n-1} \frac{1}{k}.$$
Since $\displaystyle \sum_{k=1}^{n-1} \frac{1}{k} \geq 1$ for all $n\geq 2$ so $d_n$ is indeed monotone.
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
Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:
$$S=\sum_{n=0}^\infty\frac{1}{3^n}\sum_{k=0}^n\frac{1}{3^{-k}}\frac{(-1)^{k+1}}{k+1}$$
where
$$\sum_{k=0}^n\frac{1}{3^{-k}}\frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}\Phi(-3,1,n+2)-\frac{2}{3}\log(2)$$
where $\Phi(a,b,c)$ is so-called Lerch transcendent.
Also there is one useful identity: $\Phi(-3, 1, n+2) = \frac{1}{\Gamma(1)}\int_0^\infty\frac{e^{-(n+2)t}}{1+3e^{-t}}dt$.
By substitution $u=1+3e^{-t}$ (and $\Gamma(1) = 1$), this integral leads to
$$\Phi(-3, 1, n+2) = \frac{1}{3^{n+2}}\int_1^4 \frac{(u-1)^{n+1}}{u} du.$$
Plugging it back (observe $u\in(1,4)$ for the convergence of the geometric series), we obtain: $$S=\sum_{n=0}^\infty\frac{1}{3^n}\left[(-3)^{n+1}\frac{1}{3^{n+2}}\int_1^4\frac{(u-1)^{n+1}}{u}du-\frac{2}{3}\log(2)\right]=$$ $$=\sum_{n=0}^\infty\frac{1}{3^{n+1}}\int_1^4\frac{(1-u)^{n+1}}{u}du-\log(2)=$$ $$=\int_1^4\frac{1}{u}\sum_{n=0}^{\infty}\left(\frac{1-u}{3}\right)^{n+1}du - \log(2)=$$ $$=\int_1^4\frac{1}{u}\frac{1-u}{u+2}du-\log(2)=-\frac{3}{2}\log(2)+\log(2)-\log(2)=-\frac{3}{2}\log(2).$$
So yes, indeed the sum converges to $S=-\frac{3}{2}\log(2)$.