Question

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

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

Short answer

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

Step-by-step solution

Step 1. Given Information 

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

Step 2. Explanation 

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

So, the series $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^k{\mathrm{\pi }}^{2\mathrm{k}}}{\left(2k\right)!}$is the maclaurin series for $\cos x\mathrm{at}x=\mathrm{\pi }$

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

Thus,

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