## Calculus: Early Transcendentals 8th Edition

$a_{n}$ = $\ln(2n^{2} +1)$ - $\ln (n^{2}+1)$ = $\ln2$
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}$ $\frac{2+\frac{1}{n^{2}}}{1+\frac{1}{n^{2}}}$ = $\ln\frac{2+0}{1+0}$ = $\ln2$.