Chapter 8: Q. 21 (page 692)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information

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

Step 2. Explanation

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

So, the given series is the maclaurin series for $\sin x a t x = π$

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

Thus,

$\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} π^{2 k + 1}}{(2 k + 1)!} = sinπ \sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} π^{2 k + 1}}{(2 k + 1)!} = 0$

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

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)kπ2k+1(2k+1)!

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Explanation

One App. One Place for Learning.

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.
$\sum_{k = 0}^{\infty} \frac{{(- 1)}^{k} π^{2 k + 1}}{(2 k + 1)!}$