## Multivariable Calculus, 7th Edition

$\Sigma^{\infty}_{n=1} \ln (\frac{n^{2}+1}{2n^{2}+1})$ $a_{n}=\ln (\frac{n^{2}+1}{2n^{2}+1})$ $\lim\limits_{n \to \infty} a_{n} = \lim\limits_{n \to \infty} \ln (\frac{n^{2}+1}{2n^{2}+1})$ $=\ln (\lim\limits_{n \to \infty}\frac{n^{2}+1}{2n^{2}+1})$ $=\ln (\lim\limits_{n \to \infty}\frac{1+\frac{1}{n^{2}}}{2+\frac{1}{n^{2}}})$ $= \ln (\frac{1+0}{2+0})$, 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.