Q. 17 - Calculus

Question

Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22.
$\sum _{k=0}^{\infty }\frac{{\left(-\mathrm{\pi }\right)}^k}{k!}$

Step-by-Step Solution

Answer

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

Step 1. Given Information

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

Step 2. Explanation

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

So, the given series is the maclaurin series for $e^x \mathrm{at} x=-\mathrm{\pi }$

Since, $\sum _{k=0}^{\infty }\frac{1}{k!}{e}^k=e^x$

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