Monotonicity, boundaries and convergence of the sequence $ \left\{ \frac{a^n}{n!} \right\} $. everyone. 
I have a doubt on the following question:
Let $ \left\{ \frac{a^n}{n!} \right\}, n \in \mathbb{N} $ be a sequence of real numbers, where $ a $ is a positive real number.
a) For what values of $ a $ is the sequence above monotonous? And bounded?
b) For what values of $ a $ is the sequence above convergent? Determine the limit of the sequence on that case.
How do I start this question?
Clearly 0 is a lower boundary. But is there a $ n_{max} $ after which $ a^n $ is always smaller than $ n! $ ? 
As far as the monotonicity goes, I thought about splitting the problem into three cases:
$ \bullet $ For $ 0 < a < 1 $:
In this case, knowing that if $ 0 < a < 1 $, then $ a^{n+1} < a^n $, we can assume that $ A_{n+1} $ is always smaller than $ A_n $ because the numerator is decreasing (as showed) and the denominator is obviously increasing. Thus, if $ a_{n+1} < a_n, \forall \,\, n \in \mathbb{N} $, the sequence is strictly decreasing and so it is monotonic.
$ \bullet $ For $ a = 1 $:
In this case, the sequence is $ \frac{1}{n!} $ which is always positive, has 0 as a lower boundary and 1 as an upper boundary and thus it is bounded. Also, given that in this case $ A_{n+1} \leq A_n, \forall \,\, n \in \mathbb{N} $ the sequence is also monotonic.
$ \bullet $ For $ a > 1 $:
In this case, $ a^{n+1} > a^n, \forall \,\, n \in \mathbb{N} $. However, in order to known whether $ a_{n+1} < a_n $ or $ a_{n+1} > a_n $ we have to check if $ a^n > n! $, which I do not know how to do.
Could anyone help me with this question?

Thanks for the attention.
Kind regards,
Pedro
 A: If
$a_n = \frac{a^n}{n!}
$,
then
$\frac{a_{n+1}}{a_n}
=\frac{\frac{a^{n+1}}{(n+1)!}}{\frac{a^n}{n!}}
=\frac{a}{n+1}
$.
Therefore,
if
$a < n+1
$,
then $a_n$ is decreasing.
In particular,
for any positive real $a$,
$a_n$ is eventually
monotonically decreasing.
If $n+1 < a$,
then
$a_n$ is increasing.
Therefore
$a_n$ first increases
and then decreases,
with a peak at about
$n=a$.
Its value there
is about
(using Stirling)
$\frac{n^n}{n!}
\approx \frac{n^n}{\sqrt{2\pi n}\frac{n^n}{e^n}}
= \frac{e^n}{\sqrt{2\pi n}}
$.
Note that
if $a=n$,
then
$a_n = a_{n+1}$
since
$\frac{n^n}{n!}
=\frac{n^{n+1}}{(n+1)!}
$.
Many questions here
show that
$a_n \to 0$
as $n \to \infty$.
You should be able to do this.
A more difficult question would be
to get a more accurate estimate
for the maximum term
depending on how close
$a$ is to an integer.
A: With the help of @marty cohen and @Victor I have conclude the following:
a)
For $ a < 1 $:
Since $ a^{n+1} < a^n $ if $ a < 1 $ and knowing also that $ \frac{1}{(n+1)!} < \frac{1}{n!}, \forall \,\, n \in \mathbb{N} $ it's easy for one to see that if $ a < 1 $, $ a_{n+1} < a_n, \forall \,\, n \in \mathbb{N} $ and thus the sequence will be strictly decreasing and as such it will be a monotonic sequence. Note also that $ \frac{a^n}{n!} > 0, \forall \,\, n \in \mathbb{N} $ and  $ a_n < a_1, \forall \,\, n \in \mathbb{N} $, so the sequence is limited.
For $ a = 1$:
In this case the sequence can be re-written as $ \frac{1}{n!} $ which is strictly decreasing also given that $ \frac{1}{(n+1)!} < \frac{1}{n!}, \forall \,\, n \in \mathbb{N} $ . The sequence is bounded inferiorly by 0 and superiorly by 1 ( the smallest value $ \frac{1}{n!} $ can assume is 1 and the sequence is strictly decreasing). Being bounded both inferiorly and superiorly, the sequence is limited.
For $ a > 1 $:
In this case the sequence does not show a constant behavior. Taking $ a = 3 $, for example - credits to @marty cohen - we have:
$ a_1 = 3, a_2 = \frac{9}{2}, a_3 = \frac{9}{2}, a_4 = \frac{81}{24} = \frac{27}{8} < \frac{9}{2} $ 
which clearly shows that the sequence first increases and then decreases. Such sequence is called a unimodal sequence and it's not monotonic.
After examining the three cases, one can conclude that the answer is $ a \leq 1 $.
b)
By using the ratio-test, one can easily see the sequence is convergent because:
$ \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left | \frac{a}{n+1} \right| = 0, \forall \,\, a \in \mathbb{R^+} $
and so the convergence of the sequence is independent of the value one gives to $ a $.
Also, looking at $ \lim_{n \to \infty} \frac{a^n}{n!} $, one can clearly see this limit results in $ 0 $. This because:
$ \bullet a^n = \underbrace{a \cdot a \cdot a \dots a}_{n \text{ times} } $
$ \bullet n! \text{ as } n \to \infty = 1 \cdot 2 \cdot 3 \dots a \cdot (a+1) \cdot (a+2) \dots (a+k) \dots n, 
 $ where $ k >> a $, and so, regardless of the value one gives to $ a $, $ n! $ will eventually outgrown $ a^n $.
Thank you, @marty cohen & @Victor for your help.
Kind regards,
Pedro.
A: Hint :
$$\frac{a^n}{n!} = \frac{\underbrace{a \cdot a \cdot a... \cdot a}_{n \text{ times}}}{1\cdot 2 \cdot 3...\cdot n}$$
So for any finite positive parameter $a$ one can assess there are two numbers $n_1$ and $n_2$ such that $n_1 < a < n_2$, where both $n_1$, $n_2 \leq n$ as $n \rightarrow \infty$.  
What can one conclude henceforth? 
