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Expert-verified Found in: Page 692 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Use an appropriate Maclaurin series to find the values of the series in Exercises 17–22. $\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}}{k}{\left(\frac{1}{\mathrm{\pi }}\right)}^{k}$

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

See the step by step solution

## Step 1. Given Information

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

## Step 2. Explanation

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

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

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

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

Thus,

$\sum _{k=0}^{\infty }\frac{{\left(-1\right)}^{k}}{k}{\left(\frac{1}{\mathrm{\pi }}\right)}^{k}=-\mathrm{ln}\left(1+\frac{1}{\mathrm{\pi }}\right)$ ### Want to see more solutions like these? 