Answer
The radius of convergence is $R = 1$, and the series converges absolutely on the interval $−1 \lt x \leq 1$.
Work Step by Step
Since
\begin{aligned}
\rho &=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\
&=\lim _{n \rightarrow \infty}\left|\frac{(-1)^{n+1} x^{n+1}}{\sqrt{n^{2}+2 n+2}} \cdot \frac{\sqrt{n^{2}+1}}{(-1)^{n} x^{n}}\right| \\
&=\lim _{n \rightarrow \infty}\left|x \frac{\sqrt{n^{2}+1}}{\sqrt{n^{2}+2 n+2}}\right|\\
&=\lim _{n \rightarrow \infty}|x \sqrt{\frac{n^{2}+1}{n^{2}+2 n+2}}|\\
&=\lim _{n \rightarrow \infty}|x \sqrt{\frac{1+1 / n^{2}}{1+2 / n+2 / n^{2}}}| \\
&=|x|
\end{aligned}
Thus the radius of convergence is $R = 1$, and the series converges absolutely on the interval $−1 \lt x \lt 1$
For the endpoint $x=1,$ the series becomes $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{\sqrt{n^{2}+1}},$ which converges by the Leibniz Test.
For $x= -1$ the series becomes $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^{2}+1}},$ which diverges by the Limit Comparison Test, comparing with the $\sum_{n=1}^{\infty} \frac{1}{n}$.
