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Chapter 1: 2P (page 17)

Test the following series for convergence.
2. ${鈭�}_{n = 1}^{鈭�} \frac{{(鈭� 2)}^{n}}{n^{2}}$

Short Answer

Expert verified

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

Step by step solution

01

Given information

It is given that the series is ${鈭�}_{n = 1}^{鈭�} \frac{{(鈭� 2)}^{n}}{n^{2}}$

02

Convergence Test

According to the convergence test for alternating series: An alternating series converges if the absolute value of the terms decreases steadily to zero, that is, if $|a_{n + 1}| 鈮� |a_{n}|$ , and $\lim_{n 鈫� 鈭�} a_{n} = 0$ .

03

Apply the properties of convergence test

Let $a_{n} = \frac{{(鈭� 2)}^{n}}{n^{2}}$ .

Then $|a_{n + 1}| = \frac{{(2)}^{n + 1}}{{(n + 1)}^{2}}$ .

Apply the convergence test as $|a_{n + 1}| 鈮� |a_{n}|$ .

To check use the following method:

If $\frac{|a_{n + 1}|}{|a_{n}|} < 1$ , that means $|a_{n}| > |a_{n + 1}|$ .

And, if $\frac{|a_{n + 1}|}{|a_{n}|} > 1$ , that means $|a_{n}| < |a_{n + 1}|$ . Thus:

$\begin{matrix} \frac{|a_{n + 1}|}{|a_{n}|} = \frac{\frac{{(2)}^{n + 1}}{{(n + 1)}^{2}}}{\frac{{(2)}^{n}}{n^{2}}} \\ = \frac{n^{2} {(2)}^{n + 1}}{{(n + 1)}^{2} {(2)}^{n}} \\ = \frac{2 n^{2}}{n^{2} + 2 n + 1} \end{matrix}$

If $(2 n^{2}) 鈭� (n^{2} + 2 n + 1) < 0$ , then $\frac{2 n^{2}}{n^{2} + 2 n + 1} < 1$ .

$(2 n^{2}) 鈭� (n^{2} + 2 n + 1) = n^{2} 鈭� 2 n 鈭� 1$

For $n 鈮� 3$ , $n^{2} 鈭� 2 n 鈭� 1 > 0$ .

That means $\frac{|a_{n + 1}|}{|a_{n}|} = \frac{\frac{{(2)}^{n + 1}}{{(n + 1)}^{2}}}{\frac{{(2)}^{n}}{n^{2}}} > 1$ for $n 鈮� 3$ .

Since, the first property is not satisfied, so the series diverges.

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Most popular questions from this chapter

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

Find the Maclaurin series $\frac{1}{\sqrt{1 + s i n (x)}}$ .

$In \cos x$ Hints:Method1:Write $\cos x = 1 + (\cos x - 1) = 1 + u$ ;use the series you know for $In (1 + u)$ ;replace u by the Maclaurin series for $(\cos x - 1)$

Method2: $In \cos x =$ $- {鈭�}_{0}^{x} \tan udu$ Use the series of Example 2 in method B.

Show that $\frac{2}{\sqrt{4 - x}} = 1 + \frac{1}{8} x$ with an error less than $\frac{1}{32}$ for 0<x<1. Hint: Let x=4y , and use the theorem (14.4) .

${鈭�}_{n = 1}^{鈭�} 鈥� {(\sqrt{x^{2} + 1})}^{n} \frac{2^{n}}{3^{n} + n^{3}}$

See all solutions

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