Proving convergence/divergence via the ratio test Consider the series
$$\sum\limits_{k=1}^\infty \frac{-3^k\cdot k!}{k^k}$$
Using the ratio test, the expression $\frac{|a_{k+1}|}{|a_k|}$ is calculated as:
$$\frac{3^{k+1}\cdot (k+1)!}{(k+1)^{k+1}}\cdot \frac{k^k}{3^k\cdot k!}=\frac{3}{(k+1)^{k}}\cdot {k^k}=3\cdot \frac{k^k}{(k+1)^{k}}=3\cdot \left(\frac{k}{k+1}\right)^k$$
How to continue? 

Seems you have problem on $\lim_{k\rightarrow \infty}\frac{k}{k+1}^k$
Let
$$y=\lim_{k\rightarrow \infty}\frac{k}{k+1}^k$$
Take $\ln$ on both sides,
$$\ln y = \lim_{k\rightarrow \infty} \frac{\ln k - \ln (k+1)}{k^{-1}}$$
Use L'Hopital's rule,
$$\lim_{k\rightarrow \infty} \frac{\ln k - \ln (k+1)}{k^{-1}} = \lim_{k\rightarrow \infty} -k^2(\frac{1}{k}-\frac{1}{k+1}) = \lim_{k\rightarrow \infty} \frac{-k}{k+1} = -1$$
Therefore,
$$y = e^{-1}$$
The ratio test shows this diverges.
