a)
The ratio test states that
[tex]\lim _{n\to\infty}\mleft|\frac{a_{n+1}}{a_n}\mright|=L[/tex]
If
• L > 1, the series diverges
• L < 1, the series converges absolutely
• L = 1, inconclusive.
Then let's apply the radio test:
[tex]a_n=\frac{2n!}{2^{2n}}\rightarrow a_{n+1}=\frac{2(n+1)!}{2^{2(n+1)}}[/tex]
Then
[tex]\lim _{n\to\infty}\mleft|\frac{a_{n+1}}{a_n}\mright|=\lim _{n\to\infty}\frac{2(n+1)!}{2^{2(n+1)}}\cdot\frac{2^{2n}}{2n!}[/tex]
Now we must simplify the expression
[tex]\begin{gathered} \lim _{n\to\infty}\mleft|\frac{a_{n+1}}{a_n}\mright|=\lim _{n\to\infty}\frac{2(n+1)!}{2n!}\cdot\frac{2^{2n}}{2^{2(n+1)}} \\ \\ \lim _{n\to\infty}|\frac{a_{n+1}}{a_n}|=\lim _{n\to\infty}\frac{(n+1)!}{n!}\cdot\frac{2^{2n}}{2^{2n}\cdot2^2} \\ \\ \lim _{n\to\infty}|\frac{a_{n+1}}{a_n}|=\lim _{n\to\infty}\frac{n+1}{4} \end{gathered}[/tex]
Then, we have
[tex]\lim _{n\to\infty}\mleft|\frac{a_{n+1}}{a_n}\mright|=\lim _{n\to\infty}\frac{n+1}{4}=+\infty[/tex]
The value of r from the ratio test is
[tex]r=+\infty[/tex]
b)
If r > 1, then we can say that the series diverges.
