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

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!

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.
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 $-\infty$.
• @Ramiro Considering that $\ln(k/(k+1) = \ln(k) - \ln(k+1)$, this answer is right on target, and your comment/review is out of line. You might consider deleting it. – Bungo Apr 18 '16 at 2:36