Find the interval of convergence of the power series and check for convergence at the...

Step 1
In mathematics, a power series (in one variable) is an infinite series of the form
${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{a}_n{(x-c)}^n=a_0+a_1(x-c)+a_2{(x-c)}^2+\cdots }$
where ${\displaystyle a_n}$ represents the coefficient of the nth term and c is a constant. Power series are useful in mathematical analysis,
where they arise as Taylor series of infinitely differentiable functions. In fact, Borel's theorem implies that every power
series is the Taylor series of some smooth function.
Step 2
The interval of convergence of the power series and check for convergence at the endpoints
${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{x^{3n+1}}{(3n+1)!}}$
Step 3
A power series ${\displaystyle \sum _{n=0}^{\mathrm{\infty }}{a}_n{(x-c)}^n}$ is convergent for some values of the variable x, which include always ${\displaystyle x=c}$ (as usual, ${\displaystyle {(x-c)}^0}$ evaluates as 1 and the sum of the series is thus ${\displaystyle a_0}$ for ${\displaystyle x=c}$). The series may diverge for other values of x. If c is not the only point of convergence, then there is always a number r with ${\displaystyle 0<r\le \mathrm{\infty }}$ such that the series converges whenever ${\displaystyle |x-c|<r}$ and diverges whenever ${\displaystyle |x-c|>r}$ The number r is called the radius of convergence of the power series; in general it is given as
${\displaystyle r=\underset{n\to \mathrm{\infty }}{lim}\in f{\left|a_n\right|}^{-\frac{1}{n}}}$
or, equivalently,
${\displaystyle r^{-1}=\underset{n\to \mathrm{\infty }}{lim}\supset {\left|a_n\right|}^{\frac{1}{n}}}$
(this is the Cauchy–Hadamard theorem; for an explanation of the notation). The relation
${\displaystyle r^{-1}=\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|}$
is also satisfied, if this limit exists.
Step 4
${\displaystyle a_n=\frac{1}{(3n+1)!}}$
${\displaystyle a_{n+1}=\frac{1}{\left(3_{n+4}\right)!}}$
${\displaystyle \frac{a_{n+1}}{a_n}=\frac{\frac{1}{\left(3_{n+4}\right)!}}{\frac{1}{\left(3_{n+1}\right)!}}}$
${\displaystyle \frac{1}{r}=\underset{n\to \mathrm{\infty }}{lim}\frac{a_{n+1}}{a_n}}$
${\displaystyle =\underset{n\to \mathrm{\infty }}{lim}\frac{(3_{n+1}!\}\{\left(3_{n+4}\right)!\}}{}}$