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? 
 A: You could use $\displaystyle\lim_{k\to\infty}3\left(\frac{k}{k+1}\right)^k=\lim_{k\to\infty}3\left(\frac{1}{1+\frac{1}{k}}\right)^k=\lim_{k\to\infty}\frac{3}{\left(1+\frac{1}{k}\right)^k}=\frac{3}{e}>1$.
A: $$\lim_{k\to\infty} 3\cdot \bigg(\frac{k}{k+1}\bigg)^k = \lim_{k\to\infty} 3\cdot \bigg(1 - \frac{1}{k+1}\bigg)^k = 3\cdot \lim_{k\to\infty}\bigg(1-\frac{1}{k}\bigg)^k = 3\cdot e^{-1}$$
A: the limit is $\frac{3}{e}>1$ thus the sum doesn't converge.
A: 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.
A: $$\log L=\lim_{k\rightarrow \infty }k\log(1-\frac{1}{k+1})=\lim_{k\rightarrow \infty }\frac{\log(1-\frac{1}{k+1})}{\frac{1}{k}}$$
by using Lopital rule
$$\log L=\lim_{k\rightarrow \infty }\frac{\frac{\frac{1}{(k+1)^2}}{1-\frac{1}{k+1}}}{\frac{-1}{k^2}}=\lim_{k\rightarrow \infty }\frac{-k^2}{(k^2+1)(1-\frac{1}{k+1})}=\lim_{k\rightarrow\infty }\frac{-1}{(1+\frac{1}{k^2})(1-\frac{1}{1+k})}=-1$$
hence the limit $$\frac{3}{e}>1$$ 
