Q. 33 - 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. 

$\sum _{k=1}^{\infty }\frac{2·4·6\cdots \left(2k\right)}{1·3·5\cdots (2k-1)}$

Step-by-Step Solution

Answer

The series diverges.

Step 1. Given information.

The given series is $\sum _{k=1}^{\infty }\frac{2·4·6\cdots \left(2k\right)}{1·3·5\cdots (2k-1)}$.

Step 2. Ratio Test.

According to the series,

$$\begin{gathered} \frac{a_{k+1}}{a_k}=\frac{\frac{2·4·6\dots \left(2k\right)(2k+2)}{1·3·5·\dots ·(2k-1)(2k+1)}}{\frac{2·4·6\cdots \left(2k\right)}{1·3·5·\cdots (2k-1)}} \\ =\frac{2k+2}{2k+1} \\ \underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}=\underset{k\to \infty }{\lim }\frac{2k+2}{2k+1} \\ =\underset{k\to \infty }{\lim }\frac{k\left(2+\frac{2}{k}\right)}{k\left(2+\frac{1}{k}\right)}=1 \end{gathered}$$

Therefore, the test is inconclusive.

Step 3. Divergence Test.

Now,

$$\begin{gathered} \left\{a_k\right\}=\left\{\frac{2·4·6·\dots \left(2k\right)}{1·3·5·\dots (2k-1)}\right\} \\ \underset{k\to \infty }{\lim }{a}_k\ne 0 \\ \text{ Here, }{a}_k\text{ does not converge to }0\text{. } \end{gathered}$$

Hence, the series is divergent.