Q. 53 - Calculus

Question

Use any convergence test from Sections 7.4–7.6 to determine whether the series in Exercises 41–59 converge or diverge. Explain why each series that meets the hypotheses of the test you select does so.  

$\sum _{k=1}^{\infty }\frac{2·5·8···\left(3k-1\right)}{3·5·7···\left(2k+1\right)}$

Step-by-Step Solution

Answer

The given series diverges.

Step 1. Given Information.

The given series is $\sum _{k=1}^{\infty }\frac{2·5·8···\left(3k-1\right)}{3·5·7···\left(2k+1\right)}.$

Step 2. Determine whether the series converges or diverges.

To determine whether the series converges or diverges we will use the ratio test since the series has positive terms that meet the hypothesis of the test.  

Let the general term is $a_k=\frac{2·5·8···\left(3k-1\right)}{3·5·7···\left(2k+1\right)}.$

So, $a_{k+1}=\frac{2·5·8···\left(3k-1\right)(3k+2)}{3·5·7···\left(2k+1\right)(2k+3)}.$

By the ratio test,

$$\begin{gathered} \rho =\underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k} \\ \rho =\underset{k\to \infty }{\lim }\frac{{\displaystyle \frac{2·5·8···\left(3k-1\right)(3k+2)}{3·5·7···\left(2k+1\right)(2k+3)}}}{{\displaystyle \frac{2·5·8···\left(3k-1\right)}{3·5·7···\left(2k+1\right)}}} \\ \rho =\underset{k\to \infty }{\lim }\frac{3k+2}{2k+3} \\ \rho =\underset{k\to \infty }{\lim }\frac{k\left(3+{\displaystyle \frac{2}{k}}\right)}{k\left(2+{\displaystyle \frac{3}{k}}\right)} \\ \rho =\frac{3}{2} \end{gathered}$$

Since $\frac{3}{2}>1$ the given series diverges.