Chapter 7: Q. 58 (page 592)

In Exercises 55– 58 use the ratio test in Theorem 7.6 to analyze the monotonicity of the given sequence.
$\{\frac{{(k!)}^{2}}{(2 k)!}\}$

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information.

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

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 $a_{k} = \frac{{(k!)}^{2}}{(2 k)!} .$

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

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

According to the ratio test,

localid="1649233296833" $\frac{a_{k + 1}}{a_{k}} = \frac{\frac{{((k + 1)!)}^{2}}{(2 (k + 1))!}}{\frac{{(k!)}^{2}}{(2 k)!}} = \frac{\frac{(k + 1)! (k + 1)!}{(2 k + 2)!}}{\frac{k! k!}{(2 k)!}} = \frac{\frac{(k + 1) k! (k + 1) k!}{(2 k + 2) (2 k + 1) (2 k)!}}{\frac{k! k!}{(2 k)!}} = \frac{(k + 1) (k + 1)}{2 (k + 1) (2 k + 1)} = \frac{k + 1}{2 (2 k + 1)} = \frac{3}{2 k + 2}$

Now, $\frac{a_{k + 1}}{a_{k}} < 1 for k \geq 1 .$

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

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In Exercises 55– 58 use the ratio test in Theorem 7.6 to analyze the monotonicity of the given sequence.
$\{\frac{{(k!)}^{2}}{(2 k)!}\}$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Use the ratio test.

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In Exercises 55– 58 use the ratio test in Theorem 7.6 to analyze the monotonicity of the given sequence.k!22k!

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Use the ratio test.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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In Exercises 55– 58 use the ratio test in Theorem 7.6 to analyze the monotonicity of the given sequence.
$\{\frac{{(k!)}^{2}}{(2 k)!}\}$