Question

Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22.

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

Short answer

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

Step-by-step solution

Step 1. Given Information    

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

Step 2. Explanation    

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

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

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

Thus,

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