Q. 20 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1. Given Information

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

Step 2. Explanation

The series can be rewritten as $\sum _{k=0}^{\infty }\frac{{(-1)}^k}{k}{\left(\frac{1}{\mathrm{\pi }}\right)}^k=-\sum _{k=0}^{\infty }\frac{{(-1)}^{k+1}}{k}{\left(\frac{1}{\mathrm{\pi }}\right)}^k$

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

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

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

Thus,

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