show the sequence has the limit 0, $x_n$=$\frac{10^{3n}}{n!}$
I started with:
$10^3$=1000
$x_1$=$\frac{10^{3(1)}}{1!}$=1000
$x_2$=$\frac{10^{3(2)}}{2!}$=500000>$x_1$
$x_3$=$\frac{10^{3(3)}}{3!}$=500000000/3>$x_2$
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How do I show the limit actually does go to 0.
For all $n > 10^3$, you have $$x_n = \frac{10^{3n}}{n!} = \frac{10^3 \times 10^3 \times ... \times 10^3}{1 \times 2 \times ... \times 10^3} \times \frac{10^3 \times ... \times 10^3}{(10^3+1) \times ... \times n}$$
so $$x_n \leq \frac{10^3 \times 10^3 \times ... \times 10^3}{1 \times 2 \times ... \times 10^3} \times \left( \frac{10^3}{10^3+1} \right)^{n-10^3}$$
Now you have a geometric sequence that converges to $0$. By comparison, $(x_n)$ tends to $0$.
Well:
$$\frac{10^{3(x+1)}}{(x+1)!}\div\frac{10^{3x}}{x!}=\frac{1000}{x+1}$$
And we have that $ x >999\implies\frac{1000}{x+1}< 1$
Hint:
First of all, obviously, we can write $10^{3n} = 1000^n$. Now, for the denominator, for $n>1000$, you have
$$n! = 1\cdot 2\cdot 3\cdots 1000\cdot 1001\cdot 1002\cdots n > 1\cdot 2\cdots 1000\cdot 1001^{n-1000} = C\cdot 1001^{n-1000}$$ for some constant $C$ (it doesn't really matter what the value of $C$ is). Can you continue from here?
You may consider first the quotients $$\frac{x_{n+1}}{x_n}= \frac{1000}{n+1} \stackrel{n \to \infty}{\longrightarrow} 0$$
So, for $N$ large enought you have vor all $n\geq N: \frac{x_{n+1}}{x_n} < \frac{1}{2}$
Now, write for $n =N+ k > N$ a telescoping product: $$x_n = \frac{x_{N+k}}{x_{N+k-1}}\cdot\frac{x_{N+k-1}}{x_{N+k-2}}\cdots \frac{x_{N+1}}{x_{N}}\cdot x_N < \frac{1}{2^{n-N}}x_N \stackrel{n \to \infty}{\longrightarrow} 0$$
The series $\sum_{n=0}^\infty x_n $ converges to $e^{10^3}$, so the sequence of terms in the series converge to zero.
Hint:
By Stirling's approximation we have $n!\sim\sqrt{2\pi n}(n/e)^n$.
$$\lim_{n\to \infty}\dfrac{x^{3n}}{n!}=\lim_{n\to \infty}\dfrac{x^{3n}e^n}{\sqrt{2\pi n}n^n}=\lim_{n\to \infty}\left(\dfrac{ex^3}{n}\right)^n\dfrac{1}{\sqrt{2\pi n}}$$
