Question

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

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

Short 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-by-step solution

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 role="math" localid="1649321001787" $f\left(x\right)=\ln (1+x)\mathrm{is}\sum _{k=0}^{\infty }\frac{{(-1)}^{k+1}}{k}{\left(x\right)}^k$

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

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

Thus,

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