Answer
Converges
Work Step by Step
Consider $a_n=\ln (1+\dfrac{1}{n^2})$ and $b_n=\dfrac{1}{n^2}$
Now, $\lim\limits_{n \to \infty}\dfrac{a_n}{b_n} =\lim\limits_{n \to \infty}\dfrac{\ln (1+\dfrac{1}{n^2})}{\dfrac{1}{n^2}}$
Now, apply L-Hospital's rule
Thus, we have $ =\lim\limits_{n \to \infty} \dfrac{\dfrac{-2}{n^3}-\dfrac{2}{n^5}}{\dfrac{-2}{n^3}}(\dfrac{n^3}{n^3})$
or, $\lim\limits_{n \to \infty} \dfrac{-2-\dfrac{2}{n^2}}{-2}$
or, $=1$
Hence, the series converges due to the limit comparison test.