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 $\sum 1/N_k^2=\infty$.

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 $\sum 1/N_k^2=\infty$.

(The converse is also false, and in fact this is what I did in the first, now edited away, version of this answer because I had misread the question.)

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 $\sum 1/N_k^2=\infty$.

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 $\sum 1/N_k^2=\infty$.