anGiven the infinite series:
[tex]\sum ^{\infty}_{n\mathop=1}(\frac{2n!}{2^{2n}})[/tex]
You need to remember that, by definition, given an infinite series:
[tex]\sum ^{\infty}_{n\mathop=1}a_n[/tex]
(a) The formula for applying the Ratio Test is:
[tex]\lim _{n\rightarrow\infty}\frac{|a_{n+1}|}{|a_n|}=L[/tex]
By definition:
1. If:
[tex]L<1[/tex]
The series converges.
2. If:
[tex]L>1[/tex]
Or:
[tex]L=\infty[/tex]
The series diverges.
3. If:
[tex]L=1[/tex]
The Ratio Test is inconclusive.
Therefore, you need to set up:
[tex]\lim _{n\rightarrow\infty}\frac{2(n+1)!}{2^{2(n+1)}_{}}\cdot\frac{2^{2n}_{}}{2n!}[/tex]
Simplifying, you get:
[tex]\lim _{n\rightarrow\infty}\frac{2(n+1)!}{2^{2(n+1)}_{}}\cdot\frac{2^{2n}_{}}{2n!}=\frac{(n+1)!}{n!}\cdot\frac{2^{2n}}{2^{2(n+1)}}=(n+1)\cdot2^{2n-2n-2}=(n+1)\cdot2^{-2}=\infty[/tex]
(b) Notice that:
[tex]r=\infty[/tex]
Therefore, this indicates that the series diverges.
Hence, the answers are:
(a)
[tex]r=\infty[/tex]
(b) It tells that the series diverges.
