The question formulated as
If $\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+2}}{a_{n}}-1)$ converges does $\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+1}}{a_{n}}-1)$ converge?
is asking if we can reduce the counter under these assumptions:
$a_{n+1} > a_{n}$
$a_{n} \in \mathbb{N}$
$\lim\limits_{n \to \infty}\frac{a_{n+1}}{a_{n}}=1$
It seems to me that the answer is generally no, but I cannot find any decisive example that would make it not working. Even more I cannot find the condition when it does work.
Reasoning: $\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+2}}{a_{n}}-1)=$
$=\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+2}}{a_{n}}-\frac{a_{n+1}}{a_{n}}+\frac{a_{n+1}}{a_{n}}-1)$
$=\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+2}-a_{n+1}}{a_{n}})+\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+1}}{a_{n}}-1)$
Now we need this to have limit and we are done:
$\sum\limits_{n=1}^{\infty}(-1)^n (\frac{a_{n+2}-a_{n+1}}{a_{n}})$
Although it might be possible that we can prove that this is bounded, generally this can oscillate. Or?