Question

I need help solving this problem It’s from my calculus prep guide

Answer 1

Given the expression:

`\sum ^(\infty)_(n\mathop=1)((2n!)/(2^(2n)))`

a) What is the value of (r) from the ratio test?

The value of (r) will be calculated using the following formula:

`r=\lim _(n\rightarrow\infty)|(a_(n+1))/(a_n)|`

So, we will calculate it as follows:

`\begin{gathered} a_n=((2n!)/(2^(2n))) \\ a_(n+1)=((2(n+1)!)/(2^(2(n+1)))) \end{gathered}`

Substitute into the formula, then find the limit

`\begin{gathered} r=\lim _(n\rightarrow\infty)((2(n+1)!)/(2^(2(n+1)))*(2^(2n))/(2n!)) \\ \\ =\lim _(n\rightarrow\infty)((2(n+1)!)/(2n!)*(2^(2n))/(2^(2(n+1)))) \end{gathered}`

Simplify:

`\begin{gathered} r=\lim _(n\rightarrow\infty)((2(n+1)\cdot n!)/(2n!)*(2^(2n))/(2^(2n)\cdot2^2)) \\ \\ =\lim _(n\rightarrow\infty)((n+1)/(2^2))=\lim _(n\rightarrow\infty)(n+1)/(4)=\infty \end{gathered}`

So, the value of r = ∞

b) What does this (r) value tell you about the series?

As shown r = ∞ > 1

so, the series diverges