Q. 18 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

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,

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