Answer
Converges to $\dfrac{1}{2+x}$ for all $|x+1| \lt 1$ or, $-2 \lt x \lt 0$
Work Step by Step
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}$
We have a convergent geometric series with first term, $a=1$ and common ratio $r =-(x+1)^n$
$S=\dfrac{1}{1-(-(x+1)^n)}=\dfrac{1}{2+x}$
Hence, the given series converges to $\dfrac{1}{2+x}$ for all $|x+1| \lt 1$ or, $-2 \lt x \lt 0$
