Answer
$a_{n}$ = $\ln(2n^{2} +1)$ - $\ln (n^{2}+1)$ = $\ln2$
Work Step by Step
Use the logarithm quotient rule, we have $a_{n}$= $\ln\frac{(2n^{2}+1)}{n^{2}+1}$.
$\lim\limits_{n \to \infty} $$\ln\frac{(2n^{2}+1)}{n^{2}+1}$ = ln $\lim\limits_{n \to \infty}$ $\frac{2n^{2}+1}{n^{2}+1}$.
Then we divide the top and bottom by $n^{2}$.
We have $\ln$$\lim\limits_{n \to \infty}$ $\frac{2+\frac{1}{n^{2}}}{1+\frac{1}{n^{2}}}$ = $\ln\frac{2+0}{1+0}$ = $\ln2$.
