Question

Use the ratio test for absolute convergence to determine whether the series in Exercises 30–35 converge absolutely or diverge.
$\sum _{k=0}^{\infty }\frac{{(-1)}^{k}1.3.5...(2k+1)}{1.4.7...\left(\right(3k+1)}$

Short answer

The series $\sum _{k=0}^{\infty }\frac{{(-1)}^{k}1.3.5...(2k+1)}{1.4.7...\left(\right(3k+1)}$converges absolutely.

Step-by-step solution

Step 1. Given Information.

The series:

$\sum _{k=0}^{\infty }\frac{{(-1)}^{k}1.3.5...(2k+1)}{1.4.7...\left(\right(3k+1)}$

Step 2. Rewrite the series.

$a_k=\frac{{(-1)}^{k}1.3.5...(2k+1)}{1.4.7...\left(\right(3k+1)}$

Step 3. Find ak+1.

$$\begin{gathered} a_{k+1}=\frac{{(-1)}^{k+1}1.3.5...\left(2\right(k+1)+1)}{1.4.7...\left(3\right(k+1)+1)} \\ =\frac{{(-1)}^{k+1}1.3.5...(2k+3)}{1.4.7...\left(\right(3k+4)} \end{gathered}$$

Step 4. Calculate ak+1ak.

$$\begin{gathered} \left|\frac{a_{k+1}}{a_k}\right|=\left|\frac{\frac{{(-1)}^{k}1.3.5...(2k+3)}{1.4.7...\left(\right(3k+4)}}{\frac{{(-1)}^{k}1.3.5...(2k+1)}{1.4.7...\left(\right(3k+1)}}\right| \\ =\left|\frac{(-1)2k+3}{3k+4}\right| \\ =\frac{2k+3}{3k+4} \end{gathered}$$

Step 5. Take limits.

$$\begin{gathered} \underset{k\to \infty }{lim}\left|\frac{a_{k+1}}{a_k}\right|=\underset{k\to \infty }{lim}\left(\frac{(2k+3)}{3k+4}\right) \\ =\underset{k\to \infty }{lim}\left(\frac{k(2+{\displaystyle \frac{3}{k}})}{k(3+{\displaystyle \frac{4}{k}})}\right) \\ =\left(\frac{2}{3}\right) \\ =\frac{2}{3}<1 \end{gathered}$$

So by the ratio test, the series converges absolutely.