Q. 22 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1. Given Information

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

Step 2. Explanation

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

So, the given series is the maclaurin series for $\frac{1}{1-x} at x=-\frac{1}{\mathrm{\pi }}$

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

Thus,

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