Answer
a) The radius of convergence is $R=1$ and the interval of convergence is: $-1 \leq x \leq 1$.
b) The interval of absolute convergence is $-1 \lt x \lt 1$ .
c) We get the value of $x$ for which the given series converges conditionally: $x=-1$.
Work Step by Step
Apply the Ratio Test to compute the interval where the series converges absolutely and the radius of convergence of the given series.
$|\dfrac{u_{n+1}}{u_{n}}|=\lim\limits_{n \to \infty} \dfrac{x^{n+1}}{\sqrt {(n+1)^2+3}} \cdot \dfrac{\sqrt {n^2+3}}{x^n}=|x| \lim\limits_{n \to \infty} |\dfrac{\sqrt {n^2+3}}{\sqrt {n^2+2n+4}}|=|x| \lim\limits_{n \to \infty} \dfrac{n \sqrt {1+\dfrac{3}{n^2}}}{n \sqrt {1+\dfrac{2}{n}+\dfrac{4}{n^2}}}|=|x|$
Thus, the series is absolutely convergent for $|x| \lt 1 \implies -1 \lt x \lt 1$.
We conclude that the series is convergent for all $x \in R$ and satisfies $-1 \lt x \lt 1$. So, the radius of convergence is $R=3$.
In order to find the interval of convergence, firstly we need to check the behaviour of the series at the end points $x=-1$ and $x=1$.
Case -1: At $x=-1$
$\sum_{n=1}^{\infty} u_n=\sum_{n=1}^{\infty} \dfrac{(-1)^n}{\sqrt {n^2+3}}$
The series $\sum_{n=1}^{\infty} \dfrac{1}{\sqrt {n^2+3}}$ is convergent by the alternating test.
Also, $\lim\limits_{n \to \infty} \dfrac{\frac{1}{\sqrt {n^2+3}}}{1/n}=\lim\limits_{n \to \infty} \dfrac{1}{\sqrt {1+\dfrac{3}{n}}}=1 \gt 0$ . The series is divergent at $x=-1$ by the limit comparison test and this means that the series is convergent but not absolutely convergent. So, the series is conditionally convergent.
Case -2: At $x=1$
$\lim\limits_{n \to \infty} \dfrac{(1)^n}{\sqrt {n^2+3}}=\lim\limits_{n \to \infty} \dfrac{1}{\sqrt {n^2+3}}=1 \gt 0$ . The series is divergent at $x=1$ by the limit comparison test and this means that the series is neither convergent nor absolutely convergent. So, the series is conditionally convergent.
Thus, we get value of $x$ for which the given series converges conditionally: $x=-1$.
Therefore, we come to the following conclusions:
a) The radius of convergence is $R=1$ and the interval of convergence is: $-1 \leq x \leq 1$.
b) The interval of absolute convergence is $-1 \lt x \lt 1$ .
c) We get value of $x$ for which the given series converges conditionally: $x=-1$.
