How can we use Stirling's approximation, $$n!\approx\sqrt{2\pi n}(n/e)^n$$ to find the size/location of the max term in, $$\sum_{n=0}^\infty\frac{x^n}{n! }$$ for any $x>0$.
I started off by writing the series in terms of Stirling's approximation, $$y=\frac{x^n}{n!}\approx \frac{x^n}{\sqrt{2\pi n}\left(n/e\right)^n}$$ I'm guessing that when $n=1$, will be the max because $n!$ increases exponentially. Am I correct, or am I misinterpreting the question?
There is no need to use any approximation for this purpose.
For a given positive value $x$ the maximum term in the series will be when :
$$\frac{x^n}{n!}>\frac{x^{n+1}}{(n+1)!}$$
and
$$\frac{x^n}{n!}>\frac{x^{n-1}}{(n-1)!}$$
It is trivial to see that this occurs when $x>n$ and $x<n+1$
So the results is that $\,x-1<n<x$
Clearly if $x$ is between 0 and 1, each term becomes smaller and smaller, so the first term ( which is $1$ ) will be the maximum.
For negative values the sign on each term alternates and this requires some other way to define the maximum term. If you use the absolute value of each term then the result is the similar to the positive values results $|x|-1<n<|x|$.
