Question

I need help with this practice problem It asks to solve (a) and (b) Please put these separately so I know which is which ^

Answer 1

Given the series

$\sum (\frac{2n!}{2^{2n}})$

Let

$a_n=\frac{2n!}{2^{2n}}$

(a) Find r using the using the formula

$r=\lim _{n\to\infty}\frac{a_{n+1}}{a_n}$

Substitute the given values.

$\begin{gathered} r=\lim _{n\to\infty}\frac{2(n+1)!}{2^{2(n+1)}}\cdot\frac{2^{2n}}{2n!} \\ =\lim _{n\to\infty}\frac{(n+1)}{2^2}\frac{^{}}{} \\ =\infty \end{gathered}$

(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.