Question

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

Short answer

The series \(\sum_{n=1}^{\infty} \frac{(-3)^{n}}{3 \cdot 5 \cdot 7 \cdot \cdot(2 n+1)}\) is convergent by the Ratio Test.

Step-by-step solution

Identify the Series

The series given is of form \(\frac{(-3)^{n}}{3 \cdot 5 \cdot 7 \cdot \ldots \cdot (2n+1)}\).

Apply the Ratio Test

The Ratio Test states that for a series \(\sum a_n\), if there exists a number \(L < 1\) such that \(\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = L < 1\), the series is convergent. Let's see if this holds true here: We want to calculate \(\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \left|\frac{(-3)^{n+1}}{3\cdot5\cdot7\cdot\ldots\cdot(2(n+1)+1)} \cdot \frac{3\cdot5\cdot7\cdot\ldots\cdot(2n+1)}{(-3)^n}\right|\). After reduction, we find that this limit is \(\lim_{n\to\infty} |-3|/((2n + 5)) = 3 / (2n+5)\). As \(n \to \infty\), this limit is equal to 0, which is less than 1.

Draw Conclusion

Since the limit of \(|a_{n+1}/a_n|\) as \(n\) approaches infinity is less than 1, by the Ratio Test, this series is convergent.