Answer
$\Sigma ln(1+a_{n})$ is convergent.
Work Step by Step
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.
