Question

I need help with this practice problem It asks to answer (a) and (b) Please put these separately so I can see which is which

Answer 1

We have a series:

`\sum ^(\infty)_{n\mathop{=}1}a_n=\sum ^(\infty)_(n=1)((2n!)/(2^(2n)))\text{.}`

(a) The value of r from the ratio test is:

`\begin{gathered} r=\lim _(n\rightarrow\infty)|(a_(n+1))/(a_n)| \\ =\lim _(n\rightarrow\infty)((2(n+1)!)/(2^(2(n+1))))/((2n!)/(2^(2n))) \\ =\lim _(n\rightarrow\infty)\frac{2(n+1)!\cdot2^(2n)}{2n!\cdot2^(2(n+1))^{}} \\ =\lim _(n\rightarrow\infty)(2(n+1)\cdot n!\cdot2^(2n))/(2n!\cdot2^(2n)\cdot2^2) \\ =\lim _(n\rightarrow\infty)((n+1))/(2^2) \\ =\infty. \end{gathered}`

(b) Because r = ∞ > 1, we conclude that the series is divergent.

Answer

(a) r = ∞

(b) The series is divergent.