I have a doubt in this exercise:
For which value of $x \in R $ the following series converges
$ \sum_{n=1}^\infty\frac{(x)^{n}}{(2+x)^n}= \sum_{n=1}^\infty(\frac{x}{2+x})^n$
If I consider the series of the absolute values $ \sum_{n=1}^\infty|\frac{x}{2+x}|^{n}$
and the root test: $ lim_{n\rightarrow \infty}\sqrt[n]{|\frac{x}{2+x}|^{n}}= lim_{n\rightarrow \infty}|\frac{x}{2+x}|= |\frac{x}{2+x}|$
the given series is absolutely convergent for: $|\frac{x}{2+x}|>1$ or $ \forall x \in (-\infty, -2) \cup (-2,- 1)$ The solution in my book is $(-\infty, -1) $
I didn't consider the value $-2 $ because the generic term of the series doesn't make sense and I notices that it is done in another exercise, too.
If your book were to indeed suggest $-2$ is a solution, then it would be wrong: even the partial sums of the series are not defined at $x=-2$, so the series cannot be (and a fortiori not converge).
However, are you certain your book is not giving the solution to be $(-1,\infty)$ (instead of what you wrote, $(-\infty, -1)$)? The correct answer is indeed $$ (-1,\infty) $$ (for instance, check that the series does converge for $x=0$, $x=1$...)
And to back this up, note that you made a mistake at the end of the root test: you want $$\left\lvert \frac{x}{2+x}\right\rvert < 1$$ not $>1$. (Also, after that you still need to check the case $=1$ by hand, i.e. to check separately whether the series converges for $x=-1$. But this is easy.)