I Hope this might help u to read better,
assess the series
infinity
sum of ((-2)^n)/(3n+1)!
n = 0
$\displaystyle \sum_{n=1}^{\infty}\frac{(-2)^n}{(3n+1)!}$
Use the generalized ratio test,
$\displaystyle \lim_{n\to\infty}\left| \frac{(-2)^{n+1}}{(3(n+1)+1)!}\cdot \frac{(3n+1)!}{(-2)^n}\right|$
Note that,
$\displaystyle (3(n+1)+1)!=(3n+4)!=(3n+4)(3n+3)(3n+2)(3n+1)!$
Thus,
$\displaystyle \left| \frac{-2}{(3n+4)(3n+3)(3n+2)} \right|= 0$ as $\displaystyle n\to\infty$
Which is less than 1 thus the series converges.
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Also, the sequence,
$\displaystyle \left\{ \frac{(-2)^n}{(3n+1)!} \right\}$
Fits the conditions for Leibniz's alternating series test, thus it absolutely convergent.
$\displaystyle \sum_{n=0}^{\infty}\frac{n!}{e^n}$
Just analytically thinking about this, I can see that the factorial will very strongly outgrow the exponential as x gets larger. But more formally...
$\displaystyle R_{n}=\frac{(n+1)!}{e^n \times e} \times \frac{e^n}{n!}$
$\displaystyle R_{n}=\frac{n}{e}$
$\displaystyle \lim_{n \to \infty}R_{n}=\infty$
Thus this diverges.
Another way is to just take the limit of the original series.
$\displaystyle \lim_{n \to \infty} \frac{n!}{e^n}=\infty$
If this limit is anything but zero, the series diverges. If it is zero, you must use another test to help you.