Rewrite $\sum_{n=1}^k\log_3{(\frac{n+1}{n})}$ and write the formula in terms of k

Question

Rewrite $\sum_{n=1}^k\log_3{(\frac{n+1}{n})}$ and write the formula in terms of k.

I rewrote to $1+\frac{1}{n}$ and summed to get (I think) $\log_3(k+\frac{1}{n^k+k!})$ but I'm unsure if the $\log_3$ can be simplified or left as is.

Answer 1

Note that $\log_3 (\frac{n+1}{n}) = \log_3 (n+1) - \log_3 n$. Summing, all terms telescope and cancel leaving only $\log_3(k+1) - \log_3 (k) = \log_3(k+1)$.

Answer 2

$$\log_3\left(\frac{2}{1}\right)+\log_3\left(\frac{3}{2}\right)+\cdots+\log_3\left(\frac{k+1}{k}\right)=\log_3\left(\frac{2\cdot3\cdots(k+1)}{1\cdot2\cdots k}\right)=\log_3\left(k+1\right)$$