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Chapter 8: Q 48. (page 670)

Find the interval of convergence for power series: ${鈭�}_{k = 0}^{鈭�} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k}$

Short Answer

Expert verified

The interval of convergence for power series is $R$ .

Step by step solution

01

Step 1. Given information.

The given power series is ${鈭�}_{k = 0}^{鈭�} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k}$ .

02

Step 2. Find the interval of convergence.

Let us assume $b_{k} = \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k}$ therefore $b_{k + 1} = \frac{{(k + 1)}^{3}}{1.3.5 . . . . . (2 (k + 1) + 1)} {(x + 1)}^{k + 1}$

Ratio for the absolute convergence is

\lim_{k 鈫� 鈭�} |\frac{b_{k + 1}}{b_{k}}| = \lim_{k 鈫� 鈭�} |\frac{\frac{{(k + 1)}^{3}}{1.3.5 . . . . (2 (k + 1) + 1)} {(x + 1)}^{k + 1}}{\frac{k^{3}}{1.3.5 . . . . . (2 k + 1)} {(x + 1)}^{k}}| = \lim_{k 鈫� 鈭�} |\frac{1}{2 k + 3} {(\frac{k}{k + 3})}^{3} (x + 1)|

So, by ratio test of absolute convergence the series will converge when $|x + 1| < 1$ .

This implies that

$- 1 < x + 1 < 1$

So, $- 1 < x + 1$ and $x + 1 < 1$

$x > - 2$ and $x < 0$

03

Step 2. Find the interval of convergence.

Evaluate the series at $x = - 2$

${鈭�}_{k = 0}^{鈭�} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k} = {鈭�}_{k = 0}^{鈭�} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(- 2 + 1)}^{k} {= 鈭�}_{k = 0}^{鈭�} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(- 1)}^{k}$

Thus, the series will diverge.

Evaluate the series when $x = 0$ .

${鈭�}_{k = 0}^{鈭�} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(x + 1)}^{k} = {鈭�}_{k = 0}^{鈭�} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(0 + 1)}^{k} = {鈭�}_{k = 0}^{鈭�} \frac{k^{3}}{1.3.5 . . . . (2 k + 1)} {(1)}^{k}$

Thus, the series will diverge.

Therefore the value of for which the series converges is $(- 2, 0)$ .

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Most popular questions from this chapter

In exercises 59-62 concern the binomial series to find the maclaurin series for the given function .

$\sqrt[3]{1 + x}$

In Exercises 49鈥�56 find the Taylor series for the specified function and the given value of $x_{0}$ . Note: These are the same functions and values as in Exercises 41鈥�48.

$55 . \cos 2 x, \frac{蟺}{4}$

Let f be a twice-differentiable function at a point $x_{0}$ . Explain why the sum

$f (x) + f' (x) (x - x_{0}) + \frac{f'' (x)}{2!} {(x - x_{0})}^{2}$

is not the second-order Taylor polynomial for f at $x_{0}$ .

Find the interval of convergence for each power series in Exercises 21鈥�48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.

${鈭�}_{k = 1}^{鈭�} {(- 1)}^{k} \frac{\sqrt{k}}{k^{3}} {(x - \frac{3}{2})}^{k}^{}$

What is a difference between a Taylor polynomial and the Taylor series for a function f at a point $x_{0}$ ?

See all solutions

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