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Q. 21

Expert-verifiedFound in: Page 692

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.

$\sum _{k=0}^{\infty}\frac{{(-1)}^{k}{\mathrm{\pi}}^{2\mathrm{k}+1}}{(2k+1)!}$

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

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

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

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

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

Thus,

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

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