Question

Determine the convergence or divergence of the series using any appropriate test from this chapter. Identify the test used. $$ \sum_{n=1}^{\infty} \frac{3 \cdot 5 \cdot 7 \cdot \cdot(2 n+1)}{18^{n}(2 n-1) n !} $$

Short answer

The given series converges according to the ratio test.

Step-by-step solution

Simplify the Series

The series presented is \( \sum_{n=1}^{\infty} \frac{3 \cdot 5 \cdot 7 \cdot \cdot(2 n+1)}{18^{n}(2 n-1)n !} \). First, it's noteworthy that the numerators are all odd and increase by 2 and the denominators are all even multiples of 9, which are divided by the factorial of n. This leads to the expression \( \sum_{n=1}^{\infty} \frac{(2n-1)(2n+1)}{18^{n}n !} \).

Apply the Ratio Test

In the case of this series, the ratio test is appropriate due to the factorial and exponential terms. The ratio test states that if the limit as n approaches infinity of the absolute value of the ratio of the (n+1)th term to the nth term is less than 1, then the series converges. If this limit is greater than 1, the series diverges. If it equals 1, the test is inconclusive and a different test must be used. So compute \( L = \lim_{n \to \infty} \frac{a_{n+1}}{a_{n}} = \lim_{n \to \infty} \frac{(2n+3)(2n+1)}{18^{n+1}(n+1) !} \times \frac{18^{n}n !}{(2n-1)(2n+1)} \).

Evaluate the Limit

Simplify the expression and divide each component separately. It turns out the limit \( L = \lim_{n \to \infty} \frac{(2n+3)}{18(n+1)} \). Calculate this limit to decide the convergence of the series. In fact, \( \lim_{n \to \infty} \frac{(2n+3)}{18(n+1)} \) equals \( \frac{2}{18} \) which is smaller than 1.

Convergence Conclusion

Since the limit value is less than 1, according to the ratio test, the series converges. Therefore, the series \( \sum_{n=1}^{\infty} \frac{3 \cdot 5 \cdot 7 \cdot \cdot(2 n+1)}{18^{n}(2 n-1)n !} \) converges.