Chapter 8: Q. 19 (page 692)

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

Short Answer

Expert verified

The required answer is $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + 1} {(\frac{1}{\sqrt{3}})}^{2 k + 1} = \frac{π}{6}$

Step by step solution

Step 1. Given Information

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

Step 2. Explanation

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

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

Since, $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + 1} {(x)}^{2 k + 1} = \tan^{- 1} x$

Thus,

$\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + 1} {(\frac{1}{\sqrt{3}})}^{2 k + 1} = \tan^{-} (\frac{1}{\sqrt{3}}) \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{2 k + 1} {(\frac{1}{\sqrt{3}})}^{2 k + 1} = \frac{π}{6}$

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

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Explanation

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Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22. ∑k=0∞(-1)k2k+1132k+1

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Explanation

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Most popular questions from this chapter

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