Question

HLP 20 POINTS!!! Consider the infinite series defined by. \sum_(n=1)^x ((2n!)/(2^(2n)))

What is the value of r from the ratio test?

What does this r value tell you about the series?

Answer 1

Explanation:

To find the value of r from the ratio test for the infinite series:

\sum_(n=1)^∞ ((2n!)/(2^(2n)))

We can apply the ratio test, which states that if we take the limit of the absolute value of the ratio of the (n+1)th term to the nth term as n approaches infinity, and this limit is less than 1, then the series converges absolutely. If the limit is greater than 1, then the series diverges. If the limit is exactly 1, then the test is inconclusive and we need to try another test.

So let's apply the ratio test:

r = lim_(n->∞) |((2(n+1)!)/(2^(2(n+1)))) / ((2n!)/(2^(2n)))|

r = lim_(n->∞) |((2n+2)(2n+1))/(2^2)|

r = lim_(n->∞) (2n+2)(2n+1)/(4n^2)

r = lim_(n->∞) (n+1/2)(2n+1)/n^2

r = lim_(n->∞) (2n^2 + 3n + 1)/(2n^2)

r = 2 + 3/n + 1/n^2

As n approaches infinity, the last two terms in the expression approach zero, so:

r = 2

Since r is equal to 2, which is greater than 1, the series diverges by the ratio test.

Therefore, the ratio test tells us that the given series diverges.