Question

Use the Ratio Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(2 n) !}{n^{5}} $$

Short answer

The given series ∑ ((2*n)!)/(n^5) converges according to the Ratio Test.

Step-by-step solution

Using Ratio Test

The Ratio Test states that for a series ∑ a_n , the series converges if the limit as n approaches infinity of the absolute value of a_(n+1)/a_n is less than 1. If it's greater than 1, the series diverges. If it equals 1, the test is inconclusive. Let's apply this ratio test to the series ∑ ((2*n)!)/(n^5).

Calculating (a_(n+1))/(a_n)

Calculate (a_(n+1))/(a_n) as follows: \[ \frac{(2(n+1))!/(n+1)^5}{(2n)!/n^5} = \frac{(2n+2)! *(n^5)}{(2n)! *(n+1)^5} = \frac{(2n+2)*(2n+1)*(2n)! *(n^5)}{(2n)! *(n+1)^5} = \frac{(2n+2)*(2n+1)}{(n+1)^5} \]

Evaluate the limit

Now, take the limit of the ratio from step 2 as n approaches infinity (this gives the Ratio Test's limit L): \[ \lim_{n \rightarrow \infty} \frac{(2n+2)*(2n+1)}{(n+1)^5} \] Use L'Hopital's Rule 5 times since \( \frac{(2n+2)*(2n+1)}{(n+1)^5} \) is of the form ∞/∞, and we will get L = 0 since the numerator will be 0 and the denominator will be a positive number.

Concluding the Convergence or Divergence

Since the limit L = 0 < 1, according to the Ratio Test, the given series ∑ ((2*n)!)/(n^5) converges.