Why does the series $\sum_{n=1}^\infty\frac{(-1)^{n-1}n!}{10^n}$ not converge?

Question

The question:
Test the following series for convergence or divergence: $$ \frac{1!}{10}-\frac{2!}{10^2}+\frac{3!}{10^3}-\frac{4!}{10^4}+\cdots $$ My answer:
The general term is then $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}n!}{10^n} $$ and using the alternating series test $u_{n+1}<u_n$ for all $n\ge1$ and $u_n\rightarrow0$ as $n\rightarrow\infty$.

for $n=1$:$u_n=0.1$
for $n=2$:$u_n=0.02$
for $n=3$:$u_n=0.006$
for $n=4$:$u_n=0.0024$
$\cdots$

Help

Answer 1

Consider $|\frac{u_{n+1}}{u_n}|=\frac{n+1}{10}$. $\lim_{n \to \infty}|\frac{u_{n+1}}{u_n}| > 1$. So the series diverges, by ratio test.

Answer 2

The general term does not go to 0. Therefore, the sum does not converge. You can not determine if the terms $u_n$ go to 0 just by looking at a few of them, i.e., you can not determine a limit by looking at a few values. When $n$ is small, increasing by 1 makes the denominator increase by a factor of 10, but the numerator only increases by a factor of 1 or 2 or 3. But, once $n \geq 10$, now the numerator starts increasing faster than the denominator.

In this case, $\lim_{n\to \infty} u_n = \infty$. Do you see why?

Answer 3

Short answer: $\lim_{n\to\infty}\dfrac{n!}{10^n}=\infty$, so that $\dfrac{(-1)^{n-1}n!}{10^n}$ does nort converge to $0$.