Answer
$a) R=1$. Interval of Convergence: $−1 \leq x \lt 1$
$b) $ Interval of Absolute Convergence $−1 \lt x\lt 1$
$c)$ The series converges conditionally at $x = -1$
Work Step by Step
Step 1.
Use the Ratio Test $(\frac{u_{n+1}}{u_n}) $ to find the interval where the series converges absolutely.
$\lim\limits_{n\to \infty} |\frac{x^{n+1}}{\sqrt {(n+1)^2+3}} \times \frac{\sqrt {n^2+3}}{x^n}| = \lim\limits_{n \to \infty}|\frac{x\sqrt{n^2+3}}{\sqrt {(n+1)^2+3}}| = \lim\limits_{n \to \infty} |x| \times {\lim\limits_{n \to \infty } (\frac{\sqrt {n^2+3})}{\sqrt {n^2+2n+4}})^2} = |x| \times \lim\limits_{n \to \infty} \frac{n^2+3}{n^2+2n+4} = |x| \times \lim\limits_{n \to \infty} \frac{1+\frac{3}{n^2}}{1+\frac{2}{n}+\frac{4}{n^2}} = |x| \times \lim\limits_{n \to \infty} \frac{1+0}{1+0+0} = |x| \times 1 = |x| $
$|x|<1 \\ \Rightarrow -1
