<s>[INST] 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? 
 [/INST] $$\lim_{k\to\infty} 3\cdot \bigg(\frac{k}{k+1}\bigg)^k = \lim_{k\to\infty} 3\cdot \bigg(1 - \frac{1}{k+1}\bigg)^k = 3\cdot \lim_{k\to\infty}\bigg(1-\frac{1}{k}\bigg)^k = 3\cdot e^{-1}$$
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