Question

PLZ PLZ PLZ Help!!! Consider the infinite series defined by

Answer 1

Answer:

(a)

$r =\lim_{n \to \infty} |\frac{\frac{2(n+1)!}{2^{2(n+1)}}}{\frac{2n!}{2^{2n}}}| = \lim_{n \to \infty} \frac{n+1}{4} = \infty$

(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:

$\lim_{n \to \infty} |\frac{x[n+1]}{x[n]}| = L$

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 $\frac{2n!}{2^{2n}}$.

So lets plug in n+1 and make that our numerator.$\frac{2(n+1)!}{2^{2(n+1)}}$

Now lets plug in n and make that our denominator. $\frac{2n!}{2^{2n}}$

Now set up the ratio test.

$\lim_{n \to \infty} |\frac{x[n+1]}{x[n]}| = L$

$\lim_{n \to \infty} |\frac{\frac{2(n+1)!}{2^{2(n+1)}}}{\frac{2n!}{2^{2n}}}| = L$

if you simplify this, you would get (view image for the steps)

$\lim_{n \to \infty} |\frac{\frac{2(n+1)!}{2^{2(n+1)}}}{\frac{2n!}{2^{2n}}}| = \lim_{n \to \infty} \frac{n+1}{4} = \infty$

L = infinity, which is > 1, therefore this series diverges