Chapter 9, Problem 22RE

To determine

To find: The sum of the series $\frac{1}{\sqrt{3}}-\frac{1}{9\sqrt{3}}+\frac{1}{45\sqrt{3}}-\mathrm{.....}+{\left(-1\right)}^{n-1}\frac{1}{\left(2n-1\right){\left(\sqrt{3}\right)}^{2n-1}}+\mathrm{...}$ and function and at what point.

Answer to Problem 22RE

Sum: $\frac{\pi }{6}$

Function: $f\left(x\right)={\tan }^{-1}x$

Point: $x=\frac{1}{\sqrt{3}}$

Explanation of Solution

Given information:

  $\frac{1}{\sqrt{3}}-\frac{1}{9\sqrt{3}}+\frac{1}{45\sqrt{3}}-\mathrm{.....}+{\left(-1\right)}^{n-1}\frac{1}{\left(2n-1\right){\left(\sqrt{3}\right)}^{2n-1}}+\mathrm{...}$

Concept Used:

Maclaurin Series of $f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}{x}^n}$

  ${\displaystyle \sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}{x}^n}=f\left(0\right)+\frac{f\text{'}\left(0\right)}{1!}x+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2+\mathrm{...}$

Calculation:

  $\frac{1}{\sqrt{3}}-\frac{1}{9\sqrt{3}}+\frac{1}{45\sqrt{3}}-\mathrm{.....}+{\left(-1\right)}^{n-1}\frac{1}{\left(2n-1\right){\left(\sqrt{3}\right)}^{2n-1}}+\mathrm{...}$

First rewrite the series as summation form.

So, the summation form of the series is ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}\frac{1}{\left(2n-1\right){\left(\sqrt{3}\right)}^{2n-1}}}$ .

Now compare the summation with Maclaurin series and yields,

  $\begin{array}{c}{\tan }^{-1}x={\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}\frac{x^{2n-1}}{2n-1}}\\ ={\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}\frac{{\left(\frac{1}{\sqrt{3}}\right)}^{2n-1}}{\left(2n-1\right)}}\end{array}$

So, the point is at $x=\frac{1}{\sqrt{3}}$ .

The Maclaurin series of ${\tan }^{-1}x$ is same.

So, the function is $f\left(x\right)={\tan }^{-1}x$ .

Substitute $x=\frac{1}{\sqrt{3}}$ into the function and the sum of the series is $\frac{\pi }{6}$ .