Answer
The infinite series diverges to infinity.
Work Step by Step
Let us consider an infinite series $\Sigma_{n=k}^{n=\infty} a_n$. This series will converge to the sum $S$ when the sequence of its partial sums $\{S_n\}$ converge to $S$.
That is, $S=\lim\limits_{N \to \infty} S_n$ and $S=\Sigma_{n=k}^{n=\infty} a_n$.
This series will diverge when the limit does not exist and the series will diverge to infinity when the limit is infinite.
Here, we have
$\Sigma_{k=1}^{\infty} \ln (\dfrac{k+1}{k})=\lim\limits_{n \to \infty} \Sigma_{k=1}^{n} \ln (\dfrac{k+1}{k})$
or, $=\lim\limits_{n \to \infty} [\ln (\dfrac{2}{1})+\ln (\dfrac{3}{2})+\ln (\dfrac{4}{3})+.......+\ln (\dfrac{n}{n-1})+\ln (\dfrac{n+1}{n})]$
or, $=\lim\limits_{n \to \infty} \ln[ (\dfrac{2}{1}) (\dfrac{3}{2}) (\dfrac{4}{3})...... (\dfrac{n}{n-1}) (\dfrac{n+1}{n})]$
or, $=\lim\limits_{n \to \infty} \ln[ (\dfrac{2}{1}) (\dfrac{3}{2}) (\dfrac{4}{3})...... (\dfrac{n}{n-1}) (\dfrac{n+1}{n})]$
or, $=\lim\limits_{n \to \infty} \ln (n+1)$
or, $=\infty$
This implies that the infinite series diverges to infinity.