To find the radius of convergence of the series

$\sum_{k=0}^{\infty} \frac{2k! \, x^{4k}}{(4k)^k}$

we can use the ratio test, which states that for a series $\sum a_k x^k$, the radius of convergence $R$ can be found as:

$\frac{1}{R} = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right|$

For this series, let’s focus on the term:

$a_k = \frac{2k!}{(4k)^k} x^{4k}$

The key is to examine how $a_k$ behaves as $k \to \infty$ by isolating the dependency on $x$.

Step-by-Step Solution

1. Set Up the Ratio of Consecutive Terms: $\frac{a_{k+1}}{a_k} = \frac{\frac{2(k+1)!}{(4(k+1))^{k+1}} x^{4(k+1)}}{\frac{2k!}{(4k)^k} x^{4k}}$

2. Simplify the Ratio: Since this problem’s complexity suggests direct simplification may be challenging, we can alternatively consider root convergence methods due to the structure of $x^{4k}$ term.