Chapter 1: 2P (page 17)

Test the following series for convergence.
2. $\sum_{n = 1}^{\infty} \frac{{(- 2)}^{n}}{n^{2}}$

Short Answer

Expert verified

The series $\sum_{n = 1}^{\infty} \frac{{(- 2)}^{n}}{n^{2}}$ diverges.

Step by step solution

Given information

It is given that the series is $\sum_{n = 1}^{\infty} \frac{{(- 2)}^{n}}{n^{2}}$

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 $|a_{n + 1}| \leq |a_{n}|$ , and $\lim_{n \to \infty} a_{n} = 0$ .

Apply the properties of convergence test

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

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

Apply the convergence test as $|a_{n + 1}| \leq |a_{n}|$ .

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

$(2 n^{2}) - (n^{2} + 2 n + 1) = n^{2} - 2 n - 1$

For $n \geq 3$ , $n^{2} - 2 n - 1 > 0$ .

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

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

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Test the following series for convergence.
2. $\sum_{n = 1}^{\infty} \frac{{(- 2)}^{n}}{n^{2}}$

Short Answer

Step by step solution

Given information

Convergence Test

Apply the properties of convergence test

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Test the following series for convergence.2.∑n=1∞(−2)nn2

Short Answer

Step by step solution

Given information

Convergence Test

Apply the properties of convergence test

One App. One Place for Learning.

Most popular questions from this chapter

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Test the following series for convergence.
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