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

Feb 23, 2016

In fact you just need to remember that n! > > > x^n when $n \to \infty$ for every $x \in {\mathbb{R}}^{+}$

but here the demonstration

study ${\lim}_{n \to \infty} {a}_{n}^{\frac{1}{n}}$

its x/(n!)^(1/n)

the numerator don't depend of n so we study the denominator

lim_(n->oo)(n!)^(1/n)

this limit is pretty hard, by intuition you can imagine it will be $= 1$

you can rewrite

lim_(n->oo)e^(1/nlnn!)

so we need to study lim_(n->oo)1/nln(n!)

hopefully we have this awesome approximation

so we study

${\lim}_{n \to \infty} \frac{1}{n} \left(n \ln \left(n\right) - n\right)$

which is ${\lim}_{n \to \infty} \left(\ln \left(n\right) - 1\right) = \infty$

so we proved that lim_(n->oo)(n!)^(1/n) = oo

so

${\lim}_{n \to \infty} {a}_{n}^{\frac{1}{n}} = 0$ for every $x \in {\mathbb{R}}^{+}$

radius of convergence $R = \infty$

Feb 27, 2016

Also, the Maclaren power series representation for the series e^x=sum_(n=0)^oo x^n/(n!). So hence this power series converges to a function which is defined for all $x \in \mathbb{R}$. Hence the corresponding sequence x_n=(x^n)/(n!) must also converge for all $x \in \mathbb{R}$.

Alternatively, one could show that this is a Cauchy sequence since $\forall \epsilon > 0 , \exists k \in \mathbb{N} ,$ such that $m , n \ge k \implies | {x}_{m} - {x}_{n} | < \epsilon$

Hence it converges since by a theorem, any sequence $\left({x}_{n}\right) \in \mathbb{R}$ is convergent if and only if it is a Cauchy sequence.