This is false, and the reason for this is fairly general, so in fact pretty much any such statement has to be 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 $\sum (-1)^n b_n(2+b_n)$ implies that of $\sum (-1)^n b_n$.
This will not be the case if the convergence of the first sum depended on near perfect cancellations between potentially large contributions. Here's a concrete example: Let $f(x)=-1+\sqrt{1+x}$; notice that $f(x)>0$ and $f(x)(2+f(x))=x$. Now take $b_1=f(1/N)$, $b_n=f(1/N^2)$ for $n=2,4,6, \ldots , 2N$, and $b_n=0$ for the odd $n>1$ from this interval.
Define the whole sequence $b_n$ by a succession of intervals $I_k$ of this type, with $N_k\to\infty$ (to be chosen later). Then $\sum (-1)^nb_n(2+b_n)$ converges (to $0$). However, since $f(x)=x/2 - x^2/8+O(x^3)$ for small $x$, we have that $$ 8\sum_{n\in I_k} (-1)^n b_n =\frac{1}{N_k^2} + O(1/N_k^3) . $$ So $\sum (-1)^n b_n$ will diverge if we take $N_k$ such that $\sum 1/N_k^2=\infty$.