Proof of radius of convergence exponential function Suppose one is trying to find the radius of convergence of 
$$ \exp(x)=\sum\limits_{k=0}^{\infty} \frac{x^k}{k!} $$
This "proof" was given in the lecture:
$$ \begin{align} r =& \frac{1}{\limsup\limits_{n \rightarrow \infty}\sqrt
[n]{|a_n|}} \\ =& \frac{1}{\limsup\limits_{n \rightarrow \infty}\sqrt
[n]{\frac{1}{n!}}} \\
=&\lim\limits_{n \rightarrow \infty} \sqrt
[n]{n!}\\
r=& \infty
\end{align}$$
Other proofs involving the ratio test also seem to inspect the convergence of $\frac{1}{k!}$ instead of $\frac{x^k}{k!}$. Why is one allowed to substitute $\frac{1}{k!}$?
 A: The series converges if 
$$\limsup_{n}\sqrt[n]{\left|\frac{x^n}{n!}\right|}=|x|\limsup_{n}\sqrt[n]{\left|\frac{1}{n!}\right|}<1$$
This is tantamount to the condition 
$$|x|<\frac{1}{\limsup_{n}\sqrt[n]{\left|\frac{1}{n!}\right|}}$$
Since $\limsup_{n}\sqrt[n]{\left|\frac{1}{n!}\right|}=0$, then $\frac{1}{\limsup_{n}\sqrt[n]{\left|\frac{1}{n!}\right|}}=\infty$ and the series converges for all $x$.
A: You consider the following $$\lim_{k\to\infty}\left| \frac{x^{k+1}}{(k+1)!}\frac{k!}{x^k} \right| = \lim_{k\to\infty} \left|\frac{x}{k+1}\right| = |x|\lim_{k\to\infty}\frac{1}{k+1} = 0$$ as $\frac{1}{k+1}>0$ for $k\geq 1$. So the radius of convergence is $\infty$ and the interval of convergence is $\mathbb{R}$.
Maybe they misspelled something, and in the ratio test they wrote $\frac{1}{(k+1)!}$ In general anyway you want to show that the limit of the coefficients goes to zero, so that the whole limit goes to $0$ and has no dependence on $x$. So probably they just analysed the coefficients for that reason, as they already knew the answer.
