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Chapter 7: Q. 72 (page 593)

Determine whether the sequences in Exercises 63鈥�74 are monotonic or not. Also determine whether the given sequence is bounded or unbounded.
$\{\frac{(2 k)!}{{(k!)}^{2}}\}$

Short Answer

Expert verified

The given sequence is monotonic and unbounded.

Step by step solution

01

Step 1. Given Information.

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

02

Step 2. Determine whether the sequences are monotonic or not.

To determine whether the sequences are monotonic or not, we will use the ratio test.

Let the general term of the sequence is $a_{k} = \frac{(2 k)!}{{(k!)}^{2}} .$

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

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

According to the ratio test,

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

Now, $\frac{a_{k + 1}}{a_{k}} > 1 for k > 0 .$

Thus, the sequence is strictly increasing and it is monotonic.

03

Step 3. Determine whether the given sequence is bounded or unbounded.

From the given sequence, we can depict that it is bounded below by zero. Thus, there is no bounded above. Hence the sequence is unbounded.

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Most popular questions from this chapter

Explain why the integral test may be used to analyze the given series and then use the test to determine whether the series converges or diverges.

${鈭�}_{k = 4}^{鈭�} 鈥� \frac{2 k 鈭� 4}{k^{2} 鈭� 4 k + 3}$

Determine whether the series ${鈭�}_{k = 0}^{鈭�} e {(\frac{蟺}{3})}^{k}$ converges or diverges. Give the sum of the convergent series.

Determine whether the series ${鈭�}_{k = 2}^{鈭�} {(\frac{- 3}{5})}^{k}$ converges or diverges. Give the sum of the convergent series.

Prove Theorem 7.31. That is, show that if a function a is continuous, positive, and decreasing, and if the improper integral ${鈭�}_{1}^{鈭�} a (x) d x$ converges, then the nth remainder, $R_{n}$ , for the series ${鈭�}_{k = 1}^{鈭�} a (k)$ is bounded by $0 鈮� R_{n} = {鈭�}_{k = n + 1}^{鈭�} a (k) 鈮� {鈭�}_{n}^{鈭�} a (x) d x$

In Exercises 48鈥�51 find all values of p so that the series converges.

${鈭�}_{k = 1}^{鈭�} \frac{\ln (k)}{k^{p}}$

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