I'll start much as in Exodd's answer, but diverge a bit.
The series is $\sum_{n=0}^\infty(-1)^na_n,$ where
$$
a_n = \sum_{k=0}^n\frac1{(k + 1)(n - k + 1)} =
\frac1{n + 2}\sum_{k=0}^n\left(\frac1{k + 1} + \frac1{n - k + 1}\right) =
\frac{2H_{n + 1}}{n + 2},
$$
where
$$
H_m = \sum_{k=1}^m\frac1k \quad (m \geqslant 1).
$$
By the alternating series test, $\sum_{n=0}^\infty(-1)^na_n$ converges if the sequence $(a_n)$ decreases to zero (it doesn't have to be strictly decreasing) as $n$ tends to infinity. Therefore, it is enough to prove that
$$
\frac{H_m}{m + 1} \searrow 0 \text{ as } m \to \infty \quad (m \geqslant 1).
$$
It is straightforward to prove that the sequence $\left(\frac{H_m}{m + 1}\right)$ is decreasing:
\begin{gather*}
mH_m = mH_{m - 1} + 1 \leqslant (m + 1)H_{m - 1} \quad (m \geqslant 2), \\
\therefore\ \frac{H_m}{m + 1} \leqslant \frac{H_{m - 1}}m.
\end{gather*}
Here is a simple proof from first principles that $\frac{H_m}{m + 1} \to 0$ as $m \to \infty$:
\begin{gather*}
H_m \leqslant \sum_{k=1}^m\frac1{\sqrt{k}} < 2\sum_{k=1}^m\frac1{\sqrt{k} + \sqrt{k - 1}} = 2\sum_{k=1}^m\left(\sqrt{k} - \sqrt{k - 1}\right) = 2\sqrt{m}, \\
\therefore\ \frac{H_m}{m + 1} < \frac2{\sqrt{m + 1}} \to 0 \text{ as } m \to \infty.
\end{gather*}
Optional Extra
It seems a shame not to evaluate the sum, and I shall prove that it is $(\log2)^2,$ as expected.
For each finite set $F$ of ordered pairs of positive integers, define the finite sum:
$$
\mu(F) = \sum_{(i, j) \in F}\frac{(-1)^{i + j}}{ij}.
$$
For each positive integer $p,$ the $p^\text{th}$ partial sum of the given series is:
\begin{gather*}
\sum_{n=0}^{p-1}\sum_{k=0}^n\frac{(-1)^n}{(k + 1)(n - k + 1)} =
\mu(F_p), \text{ where:} \\
F_p = \{ (i, j) \colon i + j \leqslant p + 1 \}).
\end{gather*}
We have proved that $\mu(F_p)$ tends to a limit $l$ as $p \to \infty.$ Therefore:
$$
\mu(F_{4r - 1}) \to l \text{ as } r \to \infty.
$$
For each positive integer $r,$ define this "square" set of pairs of integers:
$$
K_r = \{ (i, j) \colon i \leqslant 2r \text{ and } j \leqslant 2r \}
\subset F_{4r - 1}.
$$
Then:
$$
\mu(K_r) = \left(\sum_{i=1}^{2r}\frac{(-1)^{i-1}}{i}\right)\left(\sum_{j=1}^{2r}\frac{(-1)^{j-1}}{j}\right) \to (\log2)^2 \text{ as } r \to \infty.
$$
We have $F_{4r-1} = K_r \sqcup L_r \sqcup M_r,$ a disjoint union, where:
\begin{align*}
L_r & = \{ (i, j) \colon i < 2r < j \text{ and } i + j \leqslant 4r \}, \\
M_r & = \{ (i, j) \colon j < 2r < i \text{ and } i + j \leqslant 4r \}.
\end{align*}
Clearly $\mu(M_r) = \mu(L_r),$ so we have:
$$
\mu(F_{4r-1}) = \mu(K_r) + 2\mu(L_r).
$$
We now only need to show that $\mu(L_r) \to 0$ as $r \to \infty,$ and it will follow that $l = (\log2)^2.$
\begin{align*}
\mu(L_r) & = \sum_{j=2r+1}^{4r-1}\frac{(-1)^{j-1}}{j}\sum_{i=1}^{4r-j}\frac{(-1)^{i-1}}{i} \\
& = \sum_{j=2r+1}^{4r-1}\frac{(-1)^{j-1}}{j}\sum_{i=1}^{4r-j+1}\frac{(-1)^{i-1}}{i} + \sum_{j=2r+1}^{4r-1}\frac1{j(4r-j+1)}.
\end{align*}
When $j$ is odd, $4r-j+1$ is even. When $j$ is even, $4r-j+1$ is odd. By the theory of convergent alternating series with non-zero terms, if the first term of the series is positive, the odd-numbered partial sums are greater than the infinite sum, and the even-numbered partial sums are less than the infinite sum. Hence:
\begin{align*}
\mu(L_r) & < (\log2)\!\sum_{j=2r+1}^{4r-1}\frac{(-1)^{j-1}}{j} + \frac1{4r+1}\sum_{j=2r+1}^{4r-1}\left(\frac1j + \frac1{4r-j+1}\right) \\
& = (\log2)\left(\sum_{j=1}^{4r-1}\frac{(-1)^{j-1}}{j} - \sum_{j=1}^{2r}\frac{(-1)^{j-1}}{j}\right) + \frac{H_{4r-1} - 1}{4r+1} \\
& \to 0 \text{ as } r \to \infty.
\end{align*}
The first term tends to zero because of the convergence of the series $\sum_{j=1}^\infty\frac{(-1)^{j-1}}{j},$ and the second term tends to zero because, as shown above, $H_{4r-1} < 2\sqrt{4r+1}.$ $\ \square$
Corollary.
$$
\sum_{n=1}^\infty\frac{(-1)^{n+1}H_n}{n+1} = \frac{(\log2)^2}{2}.
$$
(I've seen many similar but more complicated equations posted on Maths.SE, so this one is bound to have been posted before, probably with a much nicer proof, but it seemed worth mentioning.)
[Update]
This must be the simplest equation that comes under the heading of Euler sums. In spite of much searching, however, I haven't found it written down anywhere in such a simple form.
It is a special case of equation (2.33) in Ce Xu, Explicit evaluation of harmonic sums (2017).
Euler's equation $\sum_{n=1}^\infty\frac1{n^2} = \frac{\pi^2}6$ gives $\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2} = \frac{\pi^2}{12},$ whence:
$$
\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}n = \frac{\pi^2}{12} - \frac{(\log2)^2}{2}.
$$
Simple proofs of the latter more complex equation (it's still very simple, by the standards of results in this area!) have been requested more than once in Maths.SE:
I'm reluctant to make a fool of myself by straying ignorantly into this area of exquisite expertise. (Browse the tag just given, and you'll see what I mean.) If nevertheless it is worth polishing and simplifying my proof, I'll do so in a separate question.
Best Answer
The added terms, $-9x^3+2x^4$, will come from $c_3$ and $c_4$, in particular $a_1b_2+a_2b_1$ for $c_3$ and $a_2b_2$ for $c_4$. You have truncated the $c$ sum too early to see them.