Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

In problems 1-6, determine the convergence set of the given power series.
$\sum_{n = 0}^{\infty} \frac{n^{2}}{2^{n}} {(x + 2)}^{n}$

The set is $x \in (- 4, 0)$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1:To Find the Radius of convergence

Use The Ratio Test to find the radius of convergence. In this case,

$a_{n} = \frac{n^{2}}{2^{n}}$

Therefore,

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

and

$\begin{matrix} \lim_{n \to \infty} | \frac{a_{n}}{a_{n + 1}} | = \lim_{n \to \infty} ∣ \frac{\frac{n^{2}}{2^{n}}}{\frac{{(n + 1)}^{2}}{2^{n + 1}} ∣} \\ = \lim_{n \to \infty} | \frac{n^{2} 2^{n + 1}}{2^{n} {(n + 1)}^{2}} | \\ = \lim_{n \to \infty} | \frac{n^{2}}{{(n + 1)}^{2}} \times \frac{2^{n + 1}}{2^{n}} | \\ = \lim_{n \to \infty} | \frac{n^{2}}{{(n + 1)}^{2}} \times 2^{n + 1 - n} | \\ = \lim_{n \to \infty} | \frac{n^{2}}{{(n + 1)}^{2}} \times 2 | \end{matrix}$

Simplify further.

$\begin{matrix} \lim_{n \to \infty} | \frac{a_{n}}{a_{n + 1}} | = \lim_{n \to \infty} \frac{2 n^{2}}{n^{2} + 2 n + 1} \\ \overset{}{=} 2 \lim_{n \to \infty} \frac{n^{2}}{n^{2} (1 + \frac{2}{n} + \frac{1}{n^{2}})} \\ \overset{}{=} 2 \lim_{n \to \infty} \frac{1}{1 + \frac{2}{n} + \frac{1}{n^{2}}} \\ = 2 × \frac{1}{1} \\ = 2 \end{matrix}$

Therefore, the radius of convergence is $r = 2$ , which means that the series converges for $| x + 2 | < 2$ or, $- 2 < x + 2 < 2 \Leftrightarrow - 4 < x < 0$ that is $x \in (- 4, 0)$ .

(1): Remove the absolute value, since $\frac{n^{2}}{{(n + 1)}^{2}} > 0$ always.

(2): Strip out the largest power of in the denominator and then cancel it

(3): Here $\frac{2}{n} \to 0, n \to \infty$ and $\frac{1}{n^{2}} \to 0, n \to \infty$ .

Step 2: Check for convergence

Now, we need to check whether the boundary points $x = - 4$ and $x = 0$ are also in the set. We do that, by substituting $x = 0$ and $x = - 4$ in the series. $x = - 4$

Substituting $- 4$ for gives:

$\begin{matrix} \sum_{n = 0}^{\infty} \frac{n^{2}}{2^{n}} {(- 4 + 2)}^{n} = \sum_{n = 0}^{\infty} \frac{n^{2}}{2^{n}} {(- 2)}^{n} \\ = \sum_{n = 0}^{\infty} \frac{n^{2}}{2^{n}} \times 2^{n} \times {(- 1)}^{n} \\ = \sum_{n = 0}^{\infty} {(- 1)}^{n} n^{2} \end{matrix}$

This is an alternating series. Therefore, use the alternating series test, to check whether it converges. First, check if

$\lim_{n \to \infty} a_{n} = 0$

In this case, $a_{n} = n^{2}$ and it follows that. $\lim_{n \to \infty} n^{2} = \infty \neq 0$

(Also, $a_{n} = n^{2}$ is not monotone decreasing, which is the other condition of the alternating series test.) Therefore, the series above diverges and $x = - 4$ is not in the set.

Substituting 0 for gives:

$\begin{matrix} \sum_{n = 0}^{\infty} \frac{n^{2}}{2^{n}} {(0 + 2)}^{n} = \sum_{n = 0}^{\infty} \frac{n^{2}}{2^{n}} 2^{n} \\ = \sum_{n = 0}^{\infty} n^{2} \end{matrix}$

Here, use the fact that if a series converges, then $\lim_{n \to \infty} a_{n} = 0$

By contraposition, it follows that

$p \Rightarrow q \Leftrightarrow \neg q \Rightarrow \neg p$

In this case, that is:

$(Series converges \Rightarrow \lim_{n \to \infty} a_{n} = 0) \Leftrightarrow (\lim_{n \to \infty} a_{n} \neq 0 \Rightarrow Series diverges)$

Now, for $a_{n} = n^{2}$ it follows that

$\lim_{n \to \infty} n^{2} = \infty \neq 0$

Therefore, the series above diverges and $x = 0$ is not in the set

. $x \in (- 4, 0)$

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Short Answer

In problems 1-6, determine the convergence set of the given power series.
$\sum_{n = 0}^{\infty} \frac{n^{2}}{2^{n}} {(x + 2)}^{n}$

Step by Step Solution

Step 1:To Find the Radius of convergence

Step 2: Check for convergence

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