Question

In Exercises 55– 58 use the ratio test in Theorem 7.6 to analyze the monotonicity of the given sequence.

$\left\{\frac{3^k}{2·4·6···\left(2k\right)}\right\}$

Short answer

The given sequence is strictly decreasing for$k\ge 1.$

Step-by-step solution

Step 1. Given Information.

The given sequence is$\left\{\frac{3^k}{2·4·6···\left(2k\right)}\right\}.$

Step 2. Use the ratio test. 

To analyze the monotonicity of the given sequence we will use the ratio test.

Let the general term of the sequence is $a_k=\frac{3^k}{2·4·6···\left(2k\right)}.$

So, the term $a_{k+1}$is

$a_{k+1}=\frac{3^{k+1}}{2·4·6···\left(2k\right)\left(2\left(k+1\right)\right)}.$

According to the ratio test,

localid="1649233002767" $$\begin{gathered} \frac{a_{k+1}}{a_k} \\ =\frac{\frac{3^{k+1}}{2·4·6···\left(2k\right)\left(2\left(k+1\right)\right)}}{\frac{3^k}{2·4·6···\left(2k\right)}} \\ =\frac{3^k·3}{3^k\left(2k+2\right)} \\ =\frac{3}{2k+2} \end{gathered}$$

Now, $\frac{a_{k+1}}{a_k}<1\mathrm{for}k\ge 1.$

Thus, the given sequence is strictly decreasing for$k\ge 1.$