Question

Use the Ratio Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n}[2 \cdot 4 \cdot 6 \cdot \cdots(2 n)]}{2 \cdot 5 \cdot 8 \cdot \cdots(3 n-1)} $$

Short answer

The series converges by the Ratio Test.

Step-by-step solution

Find the General Term

The general term of the series can be expressed as: \[a_n=\frac{(-1)^n[2 \cdot 4 \cdot 6 \cdot \cdots (2n)]}{2 \cdot 5 \cdot 8 \cdot \cdots (3n-1)}\]

Compute the Ratio

The Ratio Test requires calculation of the ratio \(\)|a_{n+1}/a_n|\). Calculate the expression for \(\)|a_{n+1}/a_n|\)\[|a_{n+1}/a_n| = \left|\frac{(-1)^{n+1}(2 \cdot 4 \cdot 6 \cdots 2(n+1))}{2 \cdot 5 \cdot 8 \cdots (3(n+1)-1)} \times \frac{2 \cdot 5 \cdot 8 \cdots (3n-1)}{(-1)^n(2 \cdot 4 \cdot 6 \cdots 2n)}\right|\]\[= \left|\frac{(-1)[2(n+1)]}{[3(n+1)-1]}\right|\]\[= \frac{2(n+1)}{3n+2}\]

Compute the Limit

The next step is to find the limit of \(\)|a_{n+1}/a_n|\) as \(n\) approaches infinity, using the above expression \[ \lim_{n \to \infty} \frac{2(n+1)}{3n+2} \] Using the rules of limit, this can be simplified by dividing both the numerator and denominator by the highest power of \(n\) in the denominator. \[= \lim_{n \to \infty} \frac{2}{3} \]

Applying the Ratio Test

The Ratio Test states that if the limit of \(|a_{n+1}/a_n|\) as \(n\) approaches infinity is less than 1, the series converges; if it's more than 1, it diverges; if it's exactly 1, then the test is inconclusive. Using the result from Step 3, you see that the limit is \(2/3\), which is less than 1