The interval of convergence of a power series is by definition the set of all $x$ for which the series converges.
That the interval of convergence is, indeed, an interval follows from the following facts:
1) Suppose the power series $\sum\limits_{n=0}^\infty a_n (x-a)^n$ converges
for $x=b $. Set $|b-a|=r$. Then the power series converges
absolutely for all $x$ with $|x-a|<r$.
2) Suppose the power series $\sum\limits_{n=0}^\infty a_n (x-a)^n$ diverges for $x=b $. Set $|b-a|=r$. Then the power series diverges for all $x$ with $|x-a|>r$.
I will emphasise here: the interval of convergence of a power series $\sum\limits_{n=0}^\infty a_n (x-a)^n$ is an interval centered at $a$
Luckily, we can find "most" of the interval of convergence by first finding the
radius of convergence, $R$, of the power series. $R$ is defined to be half the length of the interval of convergence (if the interval of convergence is all of $\Bbb R$, we set $R=\infty$), and may be computed using
$$
\tag{1} R=\lim_{n\rightarrow\infty} {|a_{n}|\over |a_{n+1}|},
$$
provided this limit exists (including the case where the limit is $\infty$).
Suppose we found the radius of convergence, $R$ of a power series. Then we would know, from facts 1) and 2), that
the power series converges for any $x$ in the interval $(a-R, a+R)$, and diverges for any $x$ with $|x-a|>R$. Thus
the interval of convergence is at least the set $(a-r, a+r)\cup\{a\}$ and at most that set together with one or both of its endpoints.
Finding the interval of convergence once the radius of convergence is known is broken down into three cases:
Case 1: If $R$ is infinite, then the interval of convergence is $(-\infty,\infty)$.
Case 2: If $R=0$, then the series converges only when $x=a$, and the interval of convergence is $\{a\}$.
Case 3: When $R$ is a finite non-zero number,
the interval of convergence will be one of
$$
(a-R, a+R), [a-R, a+R), (a-R, a+R], [a-R, a+R].
$$
To determine which of the four intervals above is the interval of convergence, one must examine the series obtained
when $x$ is replaced by $a+R$ and $a-R$ separately.
So, to find the interval of convergence of the power series $\sum\limits_{n=0}^\infty a_n (x-a)^n$
1) Compute the radius of convergence, $R$, using $(1)$ (if this limit does not exist, you might try computing $R$ by using the formula given by the Root test).
2) If $R=0$ the interval of convergence is $\{a\}$.
If $R=\infty$, the interval of convergence is $(-\infty,\infty)$.
3) If $R$ is finite and non-zero, the interval of convergence is one of
$$
\tag{2} (a-R, a+R), [a-R, a+R), (a-R, a+R], [a-R, a+R].
$$
Examine the series obtained when $x=a+R$ and $x=a-R$. These will give the series
$$\sum_{n=0}^\infty a_n R^n, \text{ and }\quad\sum_{n=0}^\infty a_n(-1)^n R^n .$$ Each of these may or may not converge. You have to go back to previous methods (comparison test, alternating series test, etc.) to determine whether these series converge or not. Once you've done this, you can determine which of the four intervals in (2) is the interval of convergence.
Example:
Find the interval of convergence of the power series $\sum\limits_{n=0}^\infty
{(-1)^{n-1}\over n4^n } (x-2)^n$.
We first compute $R$:
$$\eqalign{
R&=\lim_{n\rightarrow\infty}{|a_{n}|\over |a_{n+1}|} \cr
&=\lim_{n\rightarrow\infty}{{1\over n4^n}\over{1\over (n+1)4^{n+1}}} \cr
&=\lim_{n\rightarrow\infty}{{1\over n4^n}\cdot{ (n+1)4^{n+1}}} \cr
&=\lim_{n\rightarrow\infty}{4(n+1)\over n}={4}.
}
$$
So, the radius of convergence is $R=4$. At this point, we know that the
interval of convergence is one of (using $a= 2$):
$$
[-2,6], \quad (-2,6),\quad [-2,6), \quad (-2,6].
$$
We do not know at this point what happens for $x={-2}$
or $x=-{6}$.
We must examine these cases separately.
For $x=-{ 2}$, we obtain the series
$$\sum\limits_{n=1}^\infty (-1)^{n-1} {(-2-2)^n\over n 4^n}=\sum_{n=1}^\infty\,- 1 .$$ This series diverges by the $n^{\rm th}$-Term Test. So, $x=-2$ is not in the interval of convergence.
For $x={6}$, we obtain the series
$$\sum\limits_{n=1}^\infty (-1)^{n-1} {(6-2)^n\over n 4^n}=\sum_{n=1}^\infty\,(-1)^{n-1} {1\over n } .$$ This series is a convergent Alternating series. So $x=6$ is in the interval of convergence.
The interval of convergence is thus $(-2,6]$.
Best Answer
For every $x \in \mathbb{R}$ and $n \in \mathbb{N}$ we have $$ \frac{n!x^{2n}}{n^n(1+x^{2n})} \le a_n:=\frac{n!}{n^n}, $$ and $$ \frac{a_{n+1}}{a_n}=\left(\frac{n}{n+1}\right)^n=\left(1+\frac{1}{n}\right)^{-n} \to 1/e. $$ Therefore the series converges for every $x \in \mathbb{R}$.