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Chapter 1: Q6MP (page 1)

Test for convergence: ${鈭�}_{n = 2}^{鈭�} \frac{\sqrt{n - 1}}{{(n + 1)}^{2} - 1}$

Short Answer

Expert verified

The limit is finite and the series converges.

Step by step solution

01

Concept used to prove that the series converges

Let $0 猢� a_{n} 猢� b_{n}$ for all $n$ .

Ifrole="math" localid="1664276577803" ${鈭�}_{n = 1}^{鈭�} b_{n}$ converges, then ${鈭�}_{n = 1}^{鈭�} a_{n}$ is also converges.

If ${鈭�}_{n = 1}^{鈭�} a_{n}$ diverges, then ${鈭�}_{n = 1}^{鈭�} b_{n}$ is also diverges.

02

Calculation to show that the series converges

Let the series $a_{n} = {鈭�}_{n = 2}^{鈭�} \frac{\sqrt{n - 1}}{{(n + 1)}^{2} - 1}$ .

Consider the series ${鈭�}_{n = 2}^{鈭�} \frac{\sqrt{n}}{n^{2}} = {鈭�}_{n = 2}^{鈭�} \frac{1}{n^{\frac{3}{2}}}$ .

Let $b_{n} = \frac{1}{n^{\frac{3}{2}}}$ .

The series $鈭� b_{n} = 鈭� \frac{1}{n^{\frac{3}{2}}}$ is convergent series by integral test.

Take $l i m_{n 鈫� 鈭�} \frac{a_{n}}{b_{n}}$ .

Simplify the above expression.

$l i m_{n 鈫� 鈭�} \frac{a_{n}}{b_{n}} = l i m_{n 鈫� 鈭�} (\frac{\sqrt{n - 1}}{{(n + 1)}^{2} - 1} 梅 \frac{1}{n^{\frac{3}{2}}}) = l i m_{n 鈫� 鈭�} (\frac{\sqrt{n - 1}}{{(n + 1)}^{2} - 1} 路 n^{\frac{3}{2}}) = l i m_{n 鈫� 鈭�} (\frac{\sqrt{n} \sqrt{1 - \frac{1}{n}}}{n^{2} ({(1 + \frac{1}{n})}^{2} - \frac{1}{n^{2}})} 路 n^{\frac{3}{2}})$

Solve the limit.

width="244">limn鈫�鈭�anbn=1-1w1+1w2-1(s)2

Hence, the limit is finite and the series converges.

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Most popular questions from this chapter

$\frac{1}{x} s i n x$

$f (x) = c o t x, a = \frac{蟺}{2}$

Test the following series for convergence.

2. ${鈭�}_{n = 1}^{鈭�} \frac{{(鈭� 2)}^{n}}{n^{2}}$

Show thatis the distance between the points and in the complex plane. Use this result to identify the graphs in Problems $53, 54, 59, a n d 60$ without computation.

Find the sum of each of the following series by recognizing it as the Maclaurin series for a function evaluated at a point ${鈭�}_{n = 0}^{鈭�} (\begin{matrix} - \frac{1}{2} \\ n \end{matrix}) {(- \frac{1}{2})}^{n}$ .

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