Answer:
(a)
[tex]r =\lim_{n \to \infty} |\frac{\frac{2(n+1)!}{2^{2(n+1)}}}{\frac{2n!}{2^{2n}}}| = \lim_{n \to \infty} \frac{n+1}{4} = \infty[/tex]
(b)
if r<1, then it converges absolutely
if r>1, then it diverges
if r=1, then the test is inconclusive
Step-by-step explanation:
The ratio test is:
[tex]\lim_{n \to \infty} |\frac{x[n+1]}{x[n]}| = L[/tex]
if L<1, then it converges absolutely
if L>1, then it diverges
if L=1, then the test is inconclusive
I'm assuming that they used r instead of L. Anyway, to do the ratio test, you plug in n+1 into the equation and make that numerator.
Then you plug in n into the same equation and make that your denominator.
In our case, our equation is [tex]\frac{2n!}{2^{2n}}[/tex].
So lets plug in n+1 and make that our numerator.[tex]\frac{2(n+1)!}{2^{2(n+1)}}[/tex]
Now lets plug in n and make that our denominator. [tex]\frac{2n!}{2^{2n}}[/tex]
Now set up the ratio test.
[tex]\lim_{n \to \infty} |\frac{x[n+1]}{x[n]}| = L[/tex]
[tex]\lim_{n \to \infty} |\frac{\frac{2(n+1)!}{2^{2(n+1)}}}{\frac{2n!}{2^{2n}}}| = L[/tex]
if you simplify this, you would get (view image for the steps)
[tex]\lim_{n \to \infty} |\frac{\frac{2(n+1)!}{2^{2(n+1)}}}{\frac{2n!}{2^{2n}}}| = \lim_{n \to \infty} \frac{n+1}{4} = \infty[/tex]
L = infinity, which is > 1, therefore this series diverges
