Answer
The series is absolutely convergent by root test.
Work Step by Step
$\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}(\dfrac{n^2+1}{2n^2+1})^{n}$
$|a_{n}|=(\dfrac{n^2+1}{2n^2+1})^{n}$
$\lim\limits_{n \to \infty} \sqrt[n] |a_{n}|=\lim\limits_{n \to \infty} \sqrt[n] {(\dfrac{n^2+1}{2n^2+1})^{n}}$
$=\lim\limits_{n \to \infty} \dfrac{n^2+1}{2n^2+1}$
$=\lim\limits_{n \to \infty} \dfrac{n}{2n^{2}}$
$=\frac{1}{2}\lt 1$
The series is absolutely convergent by root test.