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.
I'll not look at the provided solution, so maybe you'll get a different insight.
It's quite easy if you know about limes superior and limes inferior.
Since the given sequence is bounded, both limits (inferior and superior) are finite. There is a subsequence converging to the limes inferior and one converging to the limes superior. By assumption, these converge to the same number, therefore
$$
\liminf_{n\to\infty}a_n=\limsup_{n\to\infty}a_n
$$
and so the sequence converges.
Without the above concept, you can do as follows. First there is a converging subsequence, since the given sequence is bounded (Bolzano-Weierstraß). So we can assume, by contradiction, that the sequence doesn't converge to $a$ (the limit of all convergent subsequences).
This can mean it doesn't converge at all or that it converges to somewhere else than $a$, but it's not relevant. The important thing is that
there exists $\varepsilon>0$ such that, for all $N$, there is $n>N$ with $|a_n-a|\ge\varepsilon$.
Now choose $n_0>0$ such that $|a_{n_0}-a|\ge\varepsilon$. Next choose $n_1>n_0$ such that $|a_{n_1}-a|\ge\varepsilon$ and go on.
More precisely, if you have already chosen $n_k$ such that $n_0<n_1<\dots<n_k$ and $|a_{n_k}-a|\ge\varepsilon$, choose $n_{k+1}$ such that $|a_{n_{k+1}}-a|\ge\varepsilon$.
Thus we have a built a subsequence, but it's not necessarily convergent. However, since it is itself a bounded sequence, it has a convergent subsequence. This further subsequence cannot converge to $a$, because each of its terms $a_{n_{k_l}}$ satisfies $|a_{n_{k_l}}-a|\ge\varepsilon$. Contradiction.
Best Answer
Hint: If a series converges, then the sequence which defines the series is a null sequence.