Q. 58 - Calculus

Question

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

$\left\{\frac{{\left(k!\right)}^2}{\left(2k\right)!}\right\}$

Step-by-Step Solution

Answer

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

Step 1. Given Information.

The given sequence is $\left\{\frac{{\left(k!\right)}^2}{\left(2k\right)!}\right\}.$

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 $a_k=\frac{{\left(k!\right)}^2}{\left(2k\right)!}.$

So, the term $a_{k+1}$ is

$a_{k+1}=\frac{{\left(\left(k+1\right)!\right)}^2}{\left(2\left(k+1\right)\right)!}.$

According to the ratio test,

$$\begin{gathered} \frac{a_{k+1}}{a_k} \\ =\frac{\frac{{\left(\left(k+1\right)!\right)}^2}{\left(2\left(k+1\right)\right)!}}{\frac{{\left(k!\right)}^2}{\left(2k\right)!}} \\ =\frac{\frac{\left(k+1\right)!\left(k+1\right)!}{\left(2k+2\right)!}}{{\displaystyle \frac{k!k!}{\left(2k\right)!}}} \\ =\frac{\frac{\left(k+1\right)k!\left(k+1\right)k!}{\left(2k+2\right)\left(2k+1\right)\left(2k\right)!}}{{\displaystyle \frac{k!k!}{\left(2k\right)!}}} \\ =\frac{\left(k+1\right)\left(k+1\right)}{2\left(k+1\right)\left(2k+1\right)} \\ =\frac{k+1}{2\left(2k+1\right)} \\ =\frac{3}{2k+2} \end{gathered}$$

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

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