# Does a_n=x^n/(xn!)  converge for any x?

##### 2 Answers
Feb 6, 2017

converges for $x \in {\mathbb{R}}^{+}$

#### Explanation:

Using the Stirling asymptotic approximation

n! approx (n/e)^n we have

x^n/((x cdot n)!) approx x^n/((x cdot n)/e)^(x cdot n)=(x/x^x(e/n)^x)^n

The sequence converges if $\frac{x}{x} ^ x {\left(\frac{e}{n}\right)}^{x} < 1$ so

$\frac{e}{n} < {x}^{\frac{x - 1}{x}}$ Now given a $n$ such that $\frac{e}{n} \le \epsilon$

then the convergence is attained for ${x}^{\frac{x - 1}{x}} > \epsilon$

The function $f \left(x\right) = {x}^{\frac{x - 1}{x}}$ for $x \in {\mathbb{R}}^{+}$ has a minimum at $x = 1$ and $f \left(1\right) = 1$. Concluding the sequence is convergent for $x \in {\mathbb{R}}^{+}$

Attached a plot of $f \left(x\right)$ for $x \in {\mathbb{R}}^{+}$

Apr 19, 2017

Yes, ${\lim}_{n \to \infty} {a}_{n} = 0$ for all real numbers $x$.

#### Explanation:

Choose any real number $x$, and let $N$ be any natural number greater than or equal to $| x |$.

Let us look at $| {a}_{n} |$.

|a_n|=|x^n/(x n!)|=|x|^(n-1)/(n!) leq (N^(n-1))/(n!)

$= \frac{1}{n} \cdot \left(\frac{N}{n - 1} \cdot \frac{N}{n - 2} \cdots \frac{N}{N + 1}\right) \cdot \left(\frac{N}{N} \cdot \frac{N}{N - 1} \cdots \frac{N}{1}\right)$

leq 1/n cdot (N/N cdot N/NcdotsN/N) cdot (N^N/(N!))=1/n(N^N/(N!))

So, we have

Rightarrow 0 leq |a_n| leq 1/n(N^N/(N!)) to 0 as $n \to 0$

By Squeeze Theorem,

$R i g h t a r r o w {\lim}_{n \to \infty} | {a}_{n} | = 0$

which implies

$R i g h t a r r o w {\lim}_{n \to \infty} {a}_{n} = 0$

I hope that this was clear.