We want to get the value of r from the ratio test from the series;
[tex]\sum ^{\infty}_{n\mathop=1}(\frac{2n!}{2^{2n}})[/tex]
The ratio test is;
[tex]\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert[/tex]
We can find the limit as;
[tex]\begin{gathered} a_n=\frac{2n!}{2^{2n}} \\ a_{n+1}=\frac{2(n+1)!}{2^{2n+1}} \end{gathered}[/tex]
The quotient is; '
[tex]\frac{a_{n+1}}{a_n}=\frac{2(n+1)!(2^{2n})_{}}{2^{2(n+1)}(2n!)}=\frac{2(n+1)\times n!\times2^{2n}}{2\times2^2\times2^{2n}\times n!}=\frac{n+1}{4}[/tex]
Therefore;
[tex]r=\frac{n+1}{4}[/tex]
b. What does this value of r tell you about the series;
Taking the limit of the ratio r;
[tex]\lim _{n\to\infty}(\frac{n+1}{4})=\infty_{}[/tex]
We can tell that the series is divergent.
