Answer
The series is divergent.
Work Step by Step
$\Sigma^{\infty}_{n=1} \ln (\frac{n^{2}+1}{2n^{2}+1})$
$a_{n}=\ln (\frac{n^{2}+1}{2n^{2}+1})$
$\lim\limits_{n \to \infty} a_{n} = \lim\limits_{n \to \infty} \ln (\frac{n^{2}+1}{2n^{2}+1})$
$=\ln (\lim\limits_{n \to \infty}\frac{n^{2}+1}{2n^{2}+1})$
$=\ln (\lim\limits_{n \to \infty}\frac{1+\frac{1}{n^{2}}}{2+\frac{1}{n^{2}}})$
$= \ln (\frac{1+0}{2+0})$, since $\lim\limits_{n \to \infty} \frac{1}{n^{2}} = 0$
$= \ln (\frac{1}{2})$
$= \ln 0.5$
$\approx -0.6931$
$\ne 0$
Since $\lim\limits_{n \to \infty} a_{n} \ne 0$, then the series is divergent.