Does $\sum_{k=1}^{\infty}\ln(\frac{k}{k+1})$ converge/diverges??

Question

Does this series converge or diverge? $$\sum_{k=1}^{\infty}\ln(\frac{k}{k+1})$$

my thought is that, I can break it down to $$\sum_{k=1}^{\infty}\ln(k) - \sum_{k=1}^{\infty}\ln(k+1)$$ then maybe using comparison test or something? But I don't know exactly how to prove whether this series converges or diverges.

Any help would be appreciated!

Answer 1

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.

Answer 2

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 $-\infty$.