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Q. 20

Expert-verifiedFound in: Page 692

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.

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

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

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

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

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

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

Thus,

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

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