We have a series:
[tex]\sum ^{\infty}_{n\mathop{=}1}a_n=\sum ^{\infty}_{n=1}(\frac{2n!}{2^{2n}})\text{.}[/tex]
(a) The value of r from the ratio test is:
[tex]\begin{gathered} r=\lim _{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}| \\ =\lim _{n\rightarrow\infty}\frac{\frac{2(n+1)!}{2^{2(n+1)}}}{\frac{2n!}{2^{2n}}} \\ =\lim _{n\rightarrow\infty}\frac{2(n+1)!\cdot2^{2n}}{2n!\cdot2^{2(n+1)}^{}} \\ =\lim _{n\rightarrow\infty}\frac{2(n+1)\cdot n!\cdot2^{2n}}{2n!\cdot2^{2n}\cdot2^2} \\ =\lim _{n\rightarrow\infty}\frac{(n+1)}{2^2} \\ =\infty. \end{gathered}[/tex]
(b) Because r = ∞ > 1, we conclude that the series is divergent.
Answer
(a) r = ∞
(b) The series is divergent.
