## Calculus: Early Transcendentals 8th Edition

$\Sigma ln(1+a_{n})$ is convergent.
Given that $a_{n}\gt 0$ , we can apply the limit comparison test with $b_{n}=ln(1+a_{n})$ It is given that $\Sigma _{n=0}^{\infty}a_{n}$ is convergent. If $\lim\limits_{n\to \infty}\frac{a_{n}}{b_{n}}\ne 0$ Then according to the limit comparison test $\Sigma_{n=0}^{\infty}b_{n}$ will also converge. $\lim\limits_{n\to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n\to \infty}\frac{a_{n}}{ln(1+a_{n})}$ This is the form of $\frac{0}{0}$ we can use L-Hospital's rule. $\lim\limits_{n\to \infty}\frac{a'_{n}}{a'_{n}/(1+a'_{n})}=\lim\limits_{n\to \infty}1+a_{n}=1\ne 0$ Thus, the given series converges. Hence, $\Sigma ln(1+a_{n})$ is convergent.