<s>[INST] 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?
 [/INST] You can prove it as follows:
for every $\varepsilon >0$ and  $m+1>\left| a \right| $  and if $n$ is big enough then
 $$0<\left| \frac {a^n}{ n! }  \right| =\frac { \left| a \right|  }{ 1 } \cdot \frac { \left| a \right|  }{ 2 } \cdots \frac { \left| a \right|  }{ m } \cdot \frac { \left| a \right|  }{ m+1 } \cdots \frac { \left| a \right|  }{ n } <\frac { { \left| a \right|  }^m }{ m! } { \left( \frac { \left| a \right|  }{ m+1 }  \right)  }^{ n-m }<\varepsilon $$
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