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

Question

Samantha Adkison · Accepted Answer

In fact you just need to remember that #n! > > > x^n# when #n -> oo# for every #x in RR^+#

So the answer is yes

but here the demonstration

study #lim_(n->oo)a_n^(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

Stirling approximation

so we study

#lim_(n->oo)1/n(nln(n)-n)#

which is #lim_(n->oo)(ln(n)-1) = oo#

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

so

# lim_(n->oo)a_n^(1/n) = 0# for every #x in RR^+#

So by Cauchy-Hadamard Theorem

radius of convergence #R = oo#

Luke Alvarez · Answer

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 RR#. Hence the corresponding sequence #x_n=(x^n)/(n!)# must also converge for all #x in RR#. Alternatively, one could show that this is a Cauchy sequence since #AA epsilon >0, EE k in NN, # such that #m,n >= k => |x_m - x_n | < epsilon # Hence it converges since by a theorem, any sequence #(x_n) in RR# is convergent if and only if it is a Cauchy sequence.

Axel Alvarez · Answer

undefined The series $a_n = \frac{x^n}{n!}$ converges for all real numbers $x$.