Question

I need help with this practice problem from my trig prep book It asks to answer (a) and (b) Please put these separately ^ so I know which is which

Answer 1

Step 1:

a

`\begin{gathered} \sum ^(\infty)_(n\mathop=0)((2n!)/(2^(2n))) \\ \\ a_n\text{ = }(2n!)/(2^(2n)) \\ a_(n+1)\text{ = }\frac{2(n\text{ + 1)!}}{2^(2(n+1))} \end{gathered}`
`\begin{gathered} r\text{ = }\lim _(n\to\infty)(a_(n+1))/(a_n) \\ r\text{ =}\lim _(n\to\infty)\text{ }\frac{2(n\text{ + 1)!}}{2^(2(n+1))}\text{ }/\text{ }(2n!)/(2^(2n)) \\ r\text{ = }\lim _(n\to\infty)\frac{2(n\text{ + 1)!}}{2^(2(n+1))}\text{ }*\text{ }(2^(2n))/(2n!) \\ r\text{ = }\lim _(n\to\infty)\frac{2(n+1)n!}{2^{2n\text{ }}*\text{ 2}}\text{ }*\text{ }\frac{2^{2n^{}}}{2n!} \\ r\text{ = }\lim _(n\to\infty)(n+1)/(2) \\ \text{as n }\rightarrow\infty\text{ , r }=\text{ }\infty \end{gathered}`

b)

`\begin{gathered} \infty\text{ > 1} \\ r\text{ > 1} \end{gathered}`

If r > 1 (including infinity), then the series is divergent