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Q. 21

Expert-verified
Found in: Page 692

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22. $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}{\mathrm{\pi }}^{2\mathrm{k}+1}}{\left(2k+1\right)!}$

The required answer is $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}{\mathrm{\pi }}^{2\mathrm{k}+1}}{\left(2k+1\right)!}=0$

See the step by step solution

## Step 1. Given Information

The given series is $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}{\mathrm{\pi }}^{2\mathrm{k}+1}}{\left(2k+1\right)!}$

## Step 2. Explanation

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

So, the given series is the maclaurin series for $\mathrm{sin}xatx=\mathrm{\pi }$

Since, $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}{\mathrm{x}}^{2\mathrm{k}+1}}{\left(2k+1\right)!}=\mathrm{sin}x$

Thus,

$\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}{\mathrm{\pi }}^{2\mathrm{k}+1}}{\left(2k+1\right)!}=\mathrm{sin\pi }\phantom{\rule{0ex}{0ex}}\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}{\mathrm{\pi }}^{2\mathrm{k}+1}}{\left(2k+1\right)!}=0$