Answer
The series is divergent.
Work Step by Step
$\Sigma^{\infty}_{n=1}a_{n} = \Sigma^{\infty}_{n=1}\ln (1+\frac{1}{n}) = \Sigma^{\infty}_{n=1} \ln (\frac{n+1}{n})$
$s_{n}=\ln2+\ln\frac{3}{2}+\ln\frac{4}{3}+\ln\frac{5}{4}+\ln\frac{6}{5}+\ln\frac{7}{6}+...+\ln\frac{n}{n-1}+\ln\frac{n+1}{n}$
$=\ln2+(\ln3-\ln2)+(\ln4-\ln3)+...+(\ln(n)-\ln(n-1))+(\ln(n+1)-\ln(n))$
$s_{n} = \ln(n+1)$
$s_{n} ā \infty$ as $nā\infty$
So {$s_{n}$} is divergent.
