Chapter 1: 3P (page 17)

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

Short Answer

Expert verified

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

Step by step solution

Given information

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

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 $a_{n} = \frac{{(- 1)}^{n}}{n^{2}}$ .

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

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

$\begin{matrix} |\frac{{(- 1)}^{n + 1}}{{(n + 1)}^{2}}| \leq |\frac{{(- 1)}^{n}}{{(n)}^{2}}| \\ \frac{1}{{(n + 1)}^{2}} < \frac{1}{n^{2}} \end{matrix}$

Since the denominator is bigger, therefore the overall term is lesser. So, the first property is satisfied.

Also, it can be observed that for $\lim_{n \to \infty} a_{n} = 0$ :

$\begin{array}{l} \lim_{n \to \infty} \frac{{(- 1)}^{n}}{{(n)}^{2}} = 0 \\ \lim_{n \to \infty} \frac{{(- 1)}^{\infty}}{{(\infty)}^{2}} \approx 0 \end{array}$

Since the denominator is approaching to infinity, so the overall term will approach to zero. The second property is also satisfied.

Therefore, it is proved that the series converges.

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

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.3.∑n=1∞(−1)nn2

Short Answer

Step by step solution

Given information

Convergence Test

Apply the properties of convergence test

One App. One Place for Learning.

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