Q. 34 - Calculus

Question

In Exercises 29–34 use the ratio test to analyze whether the given series converges or diverges. If the ratio test is inconclusive, use a different test to analyze the series. 

$\sum _{k=1}^{\infty }\frac{1·4·7\cdots (3k-2)}{2·4·6\cdots \left(2k\right)}$

Step-by-Step Solution

Answer

The series diverges.

Step 1. Given information.

The given series is $\sum _{k=1}^{\infty }\frac{1·4·7\cdots (3k-2)}{2·4·6\cdots \left(2k\right)}.$

Step 2. Ratio Test.

Now,

$$\begin{gathered} \frac{a_{k+1}}{a_k}=\frac{\frac{1·4·7·\cdots (3k-2)(3k+1)}{2·4·6·\dots \left(2k\right)(2k+2)}}{\frac{1·4·7\cdots (3k-2)}{2·4·6\dots \left(2k\right)}} \\ =\frac{3k+1}{2k} \\ \underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}=\underset{k\to \infty }{\lim }\frac{3k+1}{2k} \\ =\underset{k\to \infty }{\lim }\frac{k\left(3+\frac{1}{k}\right)}{k\left(2\right)} \\ =\frac{3}{2} \\ \text{ Since, }L>1 \end{gathered}$$

Therefore, the series diverges.