Step 1:
a
[tex]\begin{gathered} \sum ^{\infty}_{n\mathop=0}(\frac{2n!}{2^{2n}}) \\ \\ a_n\text{ = }\frac{2n!}{2^{2n}} \\ a_{n+1}\text{ = }\frac{2(n\text{ + 1)!}}{2^{2(n+1)}} \end{gathered}[/tex][tex]\begin{gathered} r\text{ = }\lim _{n\to\infty}\frac{a_{n+1}}{a_n} \\ r\text{ =}\lim _{n\to\infty}\text{ }\frac{2(n\text{ + 1)!}}{2^{2(n+1)}}\text{ }\div\text{ }\frac{2n!}{2^{2n}} \\ r\text{ = }\lim _{n\to\infty}\frac{2(n\text{ + 1)!}}{2^{2(n+1)}}\text{ }\times\text{ }\frac{2^{2n}}{2n!} \\ r\text{ = }\lim _{n\to\infty}\frac{2(n+1)n!}{2^{2n\text{ }}\times\text{ 2}}\text{ }\times\text{ }\frac{2^{2n^{}}}{2n!} \\ r\text{ = }\lim _{n\to\infty}\frac{n+1}{2} \\ \text{as n }\rightarrow\infty\text{ , r }=\text{ }\infty \end{gathered}[/tex]
b)
[tex]\begin{gathered} \infty\text{ > 1} \\ r\text{ > 1} \end{gathered}[/tex]
If r > 1 (including infinity), then the series is divergent
