Q. 20 Use an appropriate Maclaurin ser... [FREE SOLUTION]

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Chapter 8: Q. 20 (page 692)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information

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

Step 2. Explanation

The series can be rewritten as ${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{k} {(\frac{1}{蟺})}^{k} = - {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k + 1}}{k} {(\frac{1}{蟺})}^{k}$

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

So, the series $- {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k + 1}}{k} {(\frac{1}{蟺})}^{k}$ is the maclaurin series for $- \ln (1 + x) a t x = \frac{1}{蟺}$

Since, $- {鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k + 1}}{k} {(\frac{1}{蟺})}^{k} = - \ln (1 + \frac{1}{蟺})$

Thus,

${鈭�}_{k = 0}^{鈭�} \frac{{(- 1)}^{k}}{k} {(\frac{1}{蟺})}^{k} = - \ln (1 + \frac{1}{蟺})$

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91影视

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

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Explanation

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91影视

Use an appropriate Maclaurin series to find the values of the series in Exercises 17鈥�22. 鈭�k=0鈭�(-1)kk1蟺k

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Explanation

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Theoretical and Mathematical Physics

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Calculus

Study anywhere. Anytime. Across all devices.

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