Find $\lim_{n\to\infty}\frac{a^n}{n!}$ First I tried to use integration:
$$y=\lim_{n\to\infty}\frac{a^n}{n!}=\lim_{n\to\infty}\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{n}$$
$$\log y=\lim_{n\to\infty}\sum_{r=1}^n\log\frac{a}{r}$$
But I could not express it as a riemann integral. Now I am thinking about sandwich theorem.
$$\frac{a}{n!}=\frac{a}{1}\cdot\frac{a}{2}\cdot\frac{a}{3}\cdots\frac{a}{t} \cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}=\frac{a}{t!}\cdot\frac{a}{t+1}\cdot\frac{a}{t+2}\cdots\frac{a}{n}$$
Since $\frac{a}{t+1}>\frac{a}{t+2}>\frac{a}{t+1}>\cdots>\frac{a}{n}$
$$\frac{a^n}{n!}<\frac{a^t}{t!}\cdot\big(\frac{a}{t+1}\big)^{n-t}$$
since $\frac{a}{t+1}<1$, $$\lim_{n\to\infty}\big(\frac{a}{t+1}\big)^{n-t}=0$$
Hence, $$\lim_{n\to\infty}\frac{a^t}{t!}\big(\frac{a}{t+1}\big)^{n-t}=0$$
And by using sandwich theorem, $y=0$. Is this correct?
 A: This is the shortest proof
$\displaystyle  \sum  _{n=1}  ^{\infty} \frac{a^{n}}{n!}$ converges by ratio test.
Let $x_{n}=\frac{a^{n}}{n!} $
Then the convergence of $\displaystyle  \sum  _{n=1} ^{\infty }x_{n} $    implies $\lbrace x_{n} \rbrace $  converges to zero
A: What is,
$$\lim_{n \to {\infty}} {{a^n} \over {n!}}$$
Well, for large values of $n$, $n!$ can be evaluated with Stirling's Approximation.
$$\lim_{n \to {\infty}} {{a^n} \over {n!}}=\lim_{n \to {\infty}} {{a^n} \over {\sqrt{2 \pi n} \cdot (n/e)^n}}$$
$${{a^n} \over {\sqrt{2 \pi n} \cdot (n/e)^n}} = {1 \over {\sqrt{2 \pi n}}} \cdot a^n \cdot (e/n)^n = {1 \over {\sqrt{2 \pi n}}} \cdot \left({{a \cdot e } \over {n}} \right)^n$$
The value of the limit is clearly $0$ for any finite $a$,
$$\lim_{n \to {\infty}} {{a^n} \over {\sqrt{2 \pi n} \cdot (n/e)^n}} = 0$$
This is because the square root term goes to zero, and because the term inside the parentheses must be less than 1 since $a$ is finite.
Thus,
$$\lim_{n \to {\infty}} {{a^n} \over {n!}}=0$$
A: You can prove it as follows:
for every $\varepsilon >0$ and  $m+1>\left| a \right| $  and if $n$ is big enough then
 $$0<\left| \frac {a^n}{ n! }  \right| =\frac { \left| a \right|  }{ 1 } \cdot \frac { \left| a \right|  }{ 2 } \cdots \frac { \left| a \right|  }{ m } \cdot \frac { \left| a \right|  }{ m+1 } \cdots \frac { \left| a \right|  }{ n } <\frac { { \left| a \right|  }^m }{ m! } { \left( \frac { \left| a \right|  }{ m+1 }  \right)  }^{ n-m }<\varepsilon $$
A: Let $x_n=\frac{a^n}{n!}$.
$$\left|\frac{x_{n+1}}{x_n}\right|=\frac{\frac{a^{n+1}}{(n+1)!}}{\frac{a^n}{n!}}=\frac{a^{n+1}n!}{a^n (n+1)!}= \frac{a}{n+1}\underset{n\to \infty }{\longrightarrow }0$$
and thus, by $x_n\to 0$ by Ratio test.
A: You missed the lower slice of bread in your sandiwch :)
Assuming that $a>0$, you should make clear that all terms are positive, no matter how obvious seems to be.
You should also write explicitly that there is some $t\in\Bbb N$ such that $t>a$. This is called "Archimedean property" of real numbers.
This would be my proof (I insist, assuming that $a>0$):
There exists some natural $t$ such that $t>a$. Then, for $n> t$
$$0<\frac{a^n}{n!}=\frac{a^t}{t!}\frac{a^{n-t}}{(t+1)\cdots n}<\frac{a^t}{t!}\left(\frac at\right)^{n-t}$$
Since $a/t<1$, the rightmost expression tends to $0$, and hence, by the sandwich theorem
$$\frac{a^n}{n!}\to 0$$
Remark: If $a\le 0$, the limit is still $0$, but in this case you should use this fact:

If $a_n$ is a sequence of real numbers such that $\lim |a_n|=0$ then $\lim a_n=0$.

A: Here's an easy way that I'm surprised is not yet here:
$$
\frac{a^n}{n!} = \frac{\overbrace{a\cdots\cdots\cdots\cdots a}}{\underbrace{1\cdot2\cdot3\cdots\,\cdots n}} 
$$
When $n$ reaches the point of being twice as big as $a,$ then every time you increment $n$ by $1$ after that, you multiply the numerator by $a$ and the denominator by more than $2a,$ so the fraction gets multiplied by something whose absolute value is less than $1/2.$ Multiplying by something less than $1/2$ over and over again will give you a product approaching $0.$
