This is false. Let's write $b_n=\frac{a_{n+1}}{a_n}-1$, so $b_n\to 0$, and we're now asking if the convergence of $\sum (-1)^n b_n$ implies that of $\sum (-1)^n b_n(b_n+2)$. Equivalently, we want to know if $\sum (-1)^n b_n^2$ is convergent in this situation.

It is clear that this is not the case. We can have cancellations that are almost completely destroyed by taking squares. For a concrete example, suppose that $b_1=1/N$, $b_n=1/N^2$ for $n=2,4,6, \ldots, 2N$, and $b_n=0$ for the odd $1<n\le 2N$.

We now define $b$ in this way on a sequence of intervals $I_k$, with $N=N_k\to\infty$. Then $\sum (-1)^n b_n$ converges (to $0$) for any such choice of $N_k$. However,
$$
\sum_{n\in I_k} (-1)^n b_n = -\frac{1}{N_k^2} + \frac{N_k}{N_k^4} ,
$$
so $\sum (-1)^n b_n$ will diverge if we choose $N_k$ such that $\sum 1/N_k^2=\infty$.