Answer
The radius of convergence is $R = 1,$ and the series converges absolutely for $-1 \lt x \lt 1.$
Work Step by Step
Apply the ratio test:
\begin{align*}
\rho&=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|\\
&=\lim _{n \rightarrow \infty}\left|\frac{x^{2 n+3}}{3 n+4} \cdot \frac{3 n+1}{x^{2 n+1}}\right|\\
&=\lim _{n \rightarrow \infty}\left|x^{2} \frac{3 n+1}{3 n+4}\right|\\
&=\left|x^{2}\right|
\end{align*}
The radius of convergence is $R = 1,$ and the series converges absolutely for $-1 \lt x \lt 1.$
For the endpoint $x = 1$, the series becomes $\sum_{n=15}^{\infty} \frac{1}{3 n+1},$ which diverges by the Limit Comparison Test.
For $x=-1 $, the series $\sum_{n=15}^{\infty} \frac{-1} {3 n+1},$ which also diverges by the Limit Comparison Test (comparing with the divergent harmonic series).
