Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{(1)(3) ....(2n-1)}{(4^n)(2^n)n!}$
Now, $l=\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_{n}} |=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(1)(3) ....(2n-1)(2n+1)}{(4^{n+1})(2^{n+1})(n+1)!}}{\dfrac{(1)(3) ....(2n-1)}{(4^n)(2^n)n!}}|$
Thus, we have $l= \lim\limits_{n \to \infty}|\dfrac{2n+1}{(4)(2) (n+1)}|=\lim\limits_{n \to \infty}|\dfrac{2+1/n}{8+8/n}|$
So, $l=\dfrac{1}{4} \lt 1$
Hence, the series Converges by the ratio test.