How do you use the ratio test to test the convergence of the series #sum_(n=1)^oo (n!)/((2n+1)!)#?

Question

Samantha Adkison · Accepted Answer

The series converges

Let #u_n=(n!)/((2n+1)!)# Then, the ratio test is #|a_(n+1)/a_(n)|=|(((n+1)!)/((2(n+1)+1)!))/((n!)/((2n+1)!))|# #=|((n+1)!)/(n!)((2n+1)!)/((2n+3)!)|# #=|(n+1)/((2n+2)(2n+3))|# #=|1/(2(2n+3))|# #1/(2(2n+3))>0# as #n in [1, +oo)# Therefore, #lim_(n->oo)|1/(2(2n+3))|=lim_(n->oo)1/(2(2n+3))# #=0# As the limit is #<1#,by the ratio test, the series converges

Emily Ammerman · Answer

undefined To use the ratio test to test the convergence of the series $ \sum_{n=1}^{\infty} \frac{n!}{(2n+1)!} $, follow these steps:

1. Write down the general term of the series, which is $ a_n = \frac{n!}{(2n+1)!} $.
2. Apply the ratio test, which states that if $ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $ exists, then:
   - If the limit is less than 1, the series converges.
   - If the limit is greater than 1 or the limit is infinity, the series diverges.
   - If the limit is equal to 1, the ratio test is inconclusive.

3. Find $ a_{n+1} $ and $ a_n $:
   - $ a_{n+1} = \frac{(n+1)!}{(2(n+1)+1)!} = \frac{(n+1)!}{(2n+3)!} $
   - $ a_n = \frac{n!}{(2n+1)!} $

4. Take the ratio $ \frac{a_{n+1}}{a_n} $ and simplify:
   - $ \frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(2n+3)!} \cdot \frac{(2n+1)!}{n!} = \frac{n+1}{(2n+2)(2n+3)} $

5. Calculate the limit as $ n \to \infty $ of the absolute value of the ratio:
   - $ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{n+1}{(2n+2)(2n+3)} $

6. Simplify and find the limit:
   - $ = \lim_{n \to \infty} \frac{1}{4} \cdot \frac{n+1}{n+\frac{3}{2}} = \frac{1}{4} $

7. Since $ \frac{1}{4} < 1 $, the series converges by the ratio test.