Does $\sum_{n=1}^\infty \frac{(-1)^n\cdot n!}{5^n}$ converge $$\sum_{n=1}^\infty \frac{(-1)^n\cdot n!}{5^n}$$
According to ratio test, $$\sum_{n=1}^\infty \frac{n!}{5^n}$$ diverges, so I believe $$\sum_{n=1}^\infty \frac{(-1)^n\cdot n!}{5^n}$$ does not converge. Is it possible that even though the alternating series test showed non-convergence, the series can still converge?
 A: Since
$$n! = \underbrace{n(n-1)\cdots 5}_{n-4 \text{ terms}}\cdot 4 \cdot 3\cdot 2\cdot 1\geq 5^{n-4}$$
we have
$$
\left|(-1)^n \frac{n!}{5^n}\right| = \frac{n!}{5^n}  \geq \frac{1}{5^4} \not\xrightarrow[n\to\infty]{}0
$$
so the terms of the series do not converge to $0$. By the limit test, the series diverges.
A: When the ratio test shows divergence, it means that the $n$th term does not go to $0$.  So, even though the ratio test is a test for absolute convergence, its failure here means that the alternating series does not converge.
A: The general term does not go to zero: $|\dfrac{ (-1)^nn!}{5^n}|\to\infty $.  Thus it diverges.
A: By Stirling approximation, $\frac{n!}{5^n} \approx \sqrt{2\pi n}(\frac{n}{5e})^n$.
Let the $n^{th}$ term of the seuqnece $x_n = \frac{(-1)^n\cdot n!}{5^n}$, then
$\ln{|x_n|} = \ln\frac{n!}{5^n} \approx (n + \frac{1}{2})\ln{n} - n(1+\ln{5}) + \frac{1}{2}\ln{n}$, the sequence clearly diverges as $n \to \infty$.
Hence, $\lim\limits_{n \to \infty} x_n \neq 0$ and the series $\sum\limits_{n=1}^{\infty} x_n$ diverges.
