Q. 21 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

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 $\sin x at x=\mathrm{\pi }$

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

Thus,

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