Answer
Converges
Work Step by Step
Let $u_n=\ln (1+\dfrac{1}{n^2})$ and $v_n=\dfrac{1}{n^2}$
Here, $\lim\limits_{n \to \infty}\dfrac{u_n}{v_n} =\lim\limits_{n \to \infty}\dfrac{\ln (1+\dfrac{1}{n^2})}{\dfrac{1}{n^2}}=\dfrac{\infty}{\infty}$
Need to apply L-Hospital's rule.
$\implies \lim\limits_{n \to \infty} \dfrac{-2/n^3-2/n^5}{-2/n^3}(\dfrac{n^3}{n^3})=\lim\limits_{n \to \infty} \dfrac{-2-\dfrac{2}{n^2}}{-2}=\dfrac{-2}{-2}=1$
Therefore, the series converges due to the limit comparison test.