Is the sequence $(x_n)=\frac{5^n}{n!}$ convergent? Is the sequence $(x_n)=\frac{5^n}{n!}$ convergent? 
I am out of my depth with this one. I took the limit of the ratio of $\frac{a_{n+1}}{a_n}$ and found it to be zero($<1$), implying that it converges to zero, but how do we prove this with the epsilon definition
(If it indeed actually converges to $0$)?
 A: Let $a_n = \dfrac{5^n}{n!}$. If $n \ge 6$ then $$0 \le a_n = \frac{5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 \cdot 5\cdots 5 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdots (n-1) \cdot n}= \frac{625}{24} \cdot \frac 56 \cdot \frac 57 \cdots \frac{5}{n} \le \frac{3125}{24 n}$$
You can either apply a squeeze theorem or write a simple $\epsilon$ argument taking advantage of this inequality.

EDIT: Let $\epsilon > 0$ be given. Choose $N \in \mathbb N$ with the property that $N > \dfrac{3125}{24 \epsilon}$. Then $$n \ge N \implies |a_n - 0| = a_n < \frac{3125}{24 n} \le \frac{3125}{24 N} < \epsilon.$$
A: Hint:
$$1\cdot2\cdot3\cdot4\cdot5\cdot6\cdots(n-1)\cdot n>1\cdot2\cdot3\cdot4\cdot5\cdot6\cdots6\cdot6$$ so that you can relate the sequence to a geometric progression of common ratio $\dfrac56<1$.
A: Since $n! > (n/e)^n, a^n/n! < a^n/(n/e)^n = (ae/n)^n$.
This goes to zero and its sum converges for any $a > 0$.
A: $e^x=\sum_{k=0}^{\infty}\dfrac{x^k}{k!}$, convergent for $x \in \mathbb{R}$.
Hence $\lim_{k \rightarrow \infty} \dfrac{x^k}{k!}=0.$
Note: Proof of the above statement: ratio test.
A: Note that by ratio test
$$\frac{x_{n+1}}{x_n}=\frac{5^{n+1}}{(n+1)!}\frac{n!}{5^n}=\frac{5}{n+1}\to 0<1$$
indeed for $n$ sufficiently large
$$x_{n}\le x_5\cdot \left(\frac56\right)^{n-5} \to 0$$
A: Use Ratio test
$$\frac{a_{n+1}}{a_n}=\frac{5^{n+1}}{(n+1)!}\frac{n!}{5^n}=\frac{5}{n+1} $$
$$\lim_{n \to \infty}\frac{5}{n+1}\to 0$$
