$\sum_{k=2}^n \log(\frac{k}{k+1})$
So far I have: $$\sum_{k=2}^n\log(k) -\log(k+1) \\ \text{Looking at this I can see that this may be a telescoping series.} \\ \sum_{k=2}^n\log(k) -\log(k+1) = [\log(2) - \log(3)] +[\log(3) - \log(4)] + \cdots + [\log(n) - \log(n+1)] \\ \text{What is in red cancels.}\\ [\log(2) - \color{red}{\log(3)}] +[\color{red}{\log(3)} - \color{red}{\log(4)}] + \cdots + [\color{red}{\log(n)} - \log(n+1)] \\ \therefore \sum_{k=2}^n \log(\frac{k}{k+1}) = \log(2) - \log(n+1)?$$
I don't know if this is correct. I tried to check it with summation calculators and they say it diverges (which I do see why), but is this how you should go about evaluating the sum?
Let $$ S = \sum_{k=2}^n \ln \left( \frac{k}{k+1} \right) \text{.} $$ Then \begin{align*} \mathrm{e}^S &= \exp\left( \sum_{k=2}^n \ln \left( \frac{k}{k+1} \right) \right) \\ &= \prod_{k=2}^n \exp \left( \ln \left( \frac{k}{k+1} \right) \right) & [ \mathrm{e}^{A+B} = \mathrm{e}^A \cdot \mathrm{e}^B] \\ &= \prod_{k=2}^n \frac{k}{k+1} \end{align*} and you should have no trouble finishing this telescoping product. What is the denominator in the last multiplicand?