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Chapter 8: Q. 19 (page 692)

Use an appropriate Maclaurin series to find the values of the series in Exercises 17鈥�22.
${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(\frac{1}{\sqrt{3}})}^{2 k + 1}$

Short Answer

Expert verified

The required answer is ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(\frac{1}{\sqrt{3}})}^{2 k + 1} = \frac{蟺}{6}$

Step by step solution

01

Step 1. Given Information

The given series is ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(\frac{1}{\sqrt{3}})}^{2 k + 1}$

02

Step 2. Explanation

The Maclaurin series for the function $f (x) = \tan^{- 1} x is \tan^{- 1} x = {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(x)}^{2 k + 1}$

So, the given series is the Maclaurin series for $\tan^{- 1} x at x = \frac{1}{\sqrt{3}}$

Since, ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(x)}^{2 k + 1} = \tan^{- 1} x$

Thus,

${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(\frac{1}{\sqrt{3}})}^{2 k + 1} = \tan^{-} (\frac{1}{\sqrt{3}}) {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{2 k + 1} {(\frac{1}{\sqrt{3}})}^{2 k + 1} = \frac{蟺}{6}$

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Most popular questions from this chapter

In Exercises 41鈥�48 find the fourth Taylor polynomial $P_{4} (x)$ for the specified function and the given value of $x_{0}$

$46 . \sqrt[3]{x}, 1$

Prove that if the power series ${鈭�}_{k = 0}^{鈭�} 鈥� a_{k} x^{k}$ has a positive and finite radius of convergence $p$ , then the series ${鈭�}_{k = 0}^{鈭�} 鈥� a_{k} x^{2 k}$ has a radius of convergence $\sqrt{p}$ .

Find the interval of convergence for power series: ${鈭�}_{k = 1}^{鈭�} {(- 1)}^{K} \frac{1}{k^{k}} {(x - 1)}^{k}$

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.

${鈭�}_{k = 0}^{鈭�} \frac{{(k!)}^{2}}{(2 k!)} {(x + 2)}^{k}$

Show that the power series ${鈭�}_{k = 1}^{鈭�} 鈥� \frac{(鈭� 1)^{k}}{k} x^{k}$ converges conditionally when $x = 1$ and diverges when $x = - 1$ . What does this behavior tell you about the interval of convergence for the series?

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