Q. 19 - Calculus

Question

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

$\sum _{k=0}^{\infty }\frac{{(-1)}^k}{2k+1}{\left(\frac{1}{\sqrt{3}}\right)}^{2k+1}$

Step-by-Step Solution

Answer

The required answer is $\sum _{k=0}^{\infty }\frac{{(-1)}^k}{2k+1}{\left(\frac{1}{\sqrt{3}}\right)}^{2k+1}=\frac{\mathrm{\pi }}{6}$

Step 1. Given Information

The given series is $\sum _{k=0}^{\infty }\frac{{(-1)}^k}{2k+1}{\left(\frac{1}{\sqrt{3}}\right)}^{2k+1}$

Step 2. Explanation

The Maclaurin series for the function $f\left(x\right)={\tan }^{-1}x \mathrm{is} {\tan }^{-1}x=\sum _{k=0}^{\infty }\frac{{(-1)}^k}{2k+1}{\left(x\right)}^{2k+1}$

So, the given series is the Maclaurin series for ${\tan }^{-1}x \mathrm{at} x=\frac{1}{\sqrt{3}}$

Since, $\sum _{k=0}^{\infty }\frac{{(-1)}^k}{2k+1}{\left(x\right)}^{2k+1}={\tan }^{-1}x$

Thus, 

$$\begin{gathered} \sum _{k=0}^{\infty }\frac{{(-1)}^k}{2k+1}{\left(\frac{1}{\sqrt{3}}\right)}^{2k+1}={\tan }^{-}\left(\frac{1}{\sqrt{3}}\right) \\ \sum _{k=0}^{\infty }\frac{{(-1)}^k}{2k+1}{\left(\frac{1}{\sqrt{3}}\right)}^{2k+1}=\frac{\mathrm{\pi }}{6} \end{gathered}$$