Given the series
[tex]\sum (\frac{2n!}{2^{2n}})[/tex]Let
[tex]a_n=\frac{2n!}{2^{2n}}[/tex](a) Find r using the using the formula
[tex]r=\lim _{n\to\infty}\frac{a_{n+1}}{a_n}[/tex]Substitute the given values.
[tex]\begin{gathered} r=\lim _{n\to\infty}\frac{2(n+1)!}{2^{2(n+1)}}\cdot\frac{2^{2n}}{2n!} \\ =\lim _{n\to\infty}\frac{(n+1)}{2^2}\frac{^{}}{} \\ =\infty \end{gathered}[/tex](b) By Ratio test, if the value of r is greater than 1, then the series is divergent. Here r > 1. So, the series is divergent.