<s>[INST] Convergence of $ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $ The task is to find out if this series is convergent or divergent.

$$ \sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! } $$

The solution uses the ratio test and says:

$ \left.\begin{aligned} \frac { a _ { n + 1 } } { a _ { n } } & = \frac { ( n + 1 ) ! ( n + 1 ) ^ { n + 1 } ( 2 n ) ! } { ( 2 ( n + 1 ) ) ! n ! n ^ { n } } = \frac { ( n + 1 ) n ! ( n + 1 ) ( n + 1 ) ^ { n } ( 2 n ) ! } { ( 2 n + 2 ) ( 2 n + 1 ) ( 2 n ) ! n ! n ^ { n } } \\ & = \frac { ( n + 1 ) ( n + 1 ) } { ( 2 n + 2 ) ( 2 n + 1 ) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } = \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } \\ & \rightarrow \frac { 1 } { 4 } \cdot e < 1 \text { for } n \rightarrow \infty \end{aligned} \right. $

I understand every step until here

$$ \frac { \left( 1 + \frac { 1 } { n } \right) \left( 1 + \frac { 1 } { n } \right) } { \left( 2 + \frac { 2 } { n } \right) \left( 2 + \frac { 1 } { n } \right) } \cdot \left( \frac { n + 1 } { n } \right) ^ { n } $$

How can all n's on the left site become 1/n? And I understand how the left site can become $\frac{1}{4}$, but how can the right site become e in the last step?
 [/INST] $\sum _ { n = 1 } ^ { \infty } \frac { n ! n ^ { n } } { ( 2 n ) ! }
$
For region of convergence,
just replace $n!$ with
$(n/e)^n$
and then use the
n-th root test.
The additional
$\sqrt{n}$
only affects the boundaries.
In this case
(why the annoying spaces?),
$\dfrac{n!n^{n}}{(2n)!}
$
becomes
$\dfrac{(n/e)^nn^{n}}{(2n/e)^{2n}}
=\left(\dfrac{(n/e)n}{(2n/e)^{2}}\right)^n
=\left(\dfrac{n^2e^2}{4n^2e}\right)^n
=\left(\dfrac{e}{4}\right)^n
$
Since
$\dfrac{e}{4} < 1$,
the sum converges.
</s>