Does $\sum_{k=1}^{\infty}\ln(\frac{k}{k+1})$ converge/diverges??
Solution 1:
You're close to the answer. Just study the $S_n$ which is the partial sum up to $n$. Infact, $S_n = \displaystyle \sum_{k=1}^n \left(\ln k - \ln(k+1)\right) = - \ln(n+1) \to -\infty$, thus the series diveges to $-\infty$.
Solution 2:
What you broke down is wrong, because $\sum \ln k $ diverges. Instead, you can calculate the partial sum directly. Let $$ S_n=\sum_{k=1}^n \ln (k/(k+1)), $$ then it is equal to $$ S_n=-\ln(n+1). $$ Therefore, given series diverges.