Discuss the convergence or divergence of $ \sum_{n=1}^{\infty} \frac {n!} {e^{n^2}} $ I did the ratio test and showed that the  limit of $ \frac {a_{n+1}}{a_n} $  is zero as n tends to infinity So I concluded that the given series is convergent but the answer is divergent .
    I checked again and again and I got the same limit zero .  Any hint would be appreciated. Thanks in advance.
 A: Through the ratio test,
$$\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{e^{(n+1)^2}} \frac{e^{n^2}}{n!}$$
We note:
$$(n+1)! = (n+1) \cdot n!$$
$$e^{(n+1)^2} = e^{n^2 + 2n + 1} = e^{n^2} \cdot e^{2n} \cdot e$$
Thus,
$$\frac{(n+1)!}{e^{(n+1)^2}} \frac{e^{n^2}}{n!} = \frac{(n+1) \cdot n!}{e^{n^2} \cdot e^{2n} \cdot e} \frac{e^{n^2}}{n!}$$
The $e^{n^2}$ and $n!$ terms cancel to give
$$\frac{(n+1) \cdot n!}{e^{n^2} \cdot e^{2n} \cdot e} \frac{e^{n^2}}{n!} =  \frac{n+1}{e^{2n} \cdot e} =  \frac{n+1}{e^{2n+1}}$$
Then, clearly, as $n \to \infty$, the ratio of successive terms approaches $0$.
... So I guess the question is, who or what is telling you the series doesn't converge? Looks like it does, and Wolfram Alpha (https://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+infinity+of+n!+%2F+e%5E(n%5E2)) also confirms as much, though Wolfram Alpha can be iffy with some series so it's not foolproof.
Granted it's been a while since I dealt with series through the ratio test so I could easily be wrong.
A: Alternatevly You can do more efficently using root test and using Stirling Approximation
$n!=\sqrt(2\pi e)(n/e)^n$
So apply n th root test 
then 
$\frac{(\sqrt(2\pi e))^{1/n}(n/e)}{e^n}=n/e^{n+1}<1 \ \forall n$ 
SO converges
