Chapter 8: Q. 20 (page 692)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information

The given series is $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{k} {(\frac{1}{π})}^{k}$

Step 2. Explanation

The series can be rewritten as $\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{k} {(\frac{1}{π})}^{k} = - \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k + 1}}{k} {(\frac{1}{π})}^{k}$

The Maclaurin series for the function role="math" localid="1649321001787" $f (x) = \ln (1 + x) is \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k + 1}}{k} {(x)}^{k}$

So, the series $- \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k + 1}}{k} {(\frac{1}{π})}^{k}$ is the maclaurin series for $- \ln (1 + x) a t x = \frac{1}{π}$

Since, $- \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k + 1}}{k} {(\frac{1}{π})}^{k} = - \ln (1 + \frac{1}{π})$

Thus,

$\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{k} {(\frac{1}{π})}^{k} = - \ln (1 + \frac{1}{π})$

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

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)kk1πk

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.
$\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k}}{k} {(\frac{1}{π})}^{k}$