Given the expression:
[tex]\sum ^{\infty}_{n\mathop=1}(\frac{2n!}{2^{2n}})[/tex]
a) What is the value of (r) from the ratio test?
The value of (r) will be calculated using the following formula:
[tex]r=\lim _{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|[/tex]
So, we will calculate it as follows:
[tex]\begin{gathered} a_n=(\frac{2n!}{2^{2n}}) \\ a_{n+1}=(\frac{2(n+1)!}{2^{2(n+1)}}) \end{gathered}[/tex]
Substitute into the formula, then find the limit
[tex]\begin{gathered} r=\lim _{n\rightarrow\infty}(\frac{2(n+1)!}{2^{2(n+1)}}\times\frac{2^{2n}}{2n!}) \\ \\ =\lim _{n\rightarrow\infty}(\frac{2(n+1)!}{2n!}\times\frac{2^{2n}}{2^{2(n+1)}}) \end{gathered}[/tex]
Simplify:
[tex]\begin{gathered} r=\lim _{n\rightarrow\infty}(\frac{2(n+1)\cdot n!}{2n!}\times\frac{2^{2n}}{2^{2n}\cdot2^2}) \\ \\ =\lim _{n\rightarrow\infty}(\frac{n+1}{2^2})=\lim _{n\rightarrow\infty}\frac{n+1}{4}=\infty \end{gathered}[/tex]
So, the value of r = ∞
b) What does this (r) value tell you about the series?
As shown r = ∞ > 1
so, the series diverges
