Question

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

Answer 1

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`

Answer 2

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.