Chapter 9, Problem 17RE

To determine

To find: The sum of the series $1-\frac{1}{4}+\frac{1}{16}-\mathrm{...}+{\left(-1\right)}^n\frac{1}{4^n}+\mathrm{...}$ and function and at what point.

Answer to Problem 17RE

Sum: $\frac{4}{5}$

Function: $f\left(x\right)=\frac{1}{1+x}$

Point: $x=\frac{1}{4}$

Explanation of Solution

Given information:

  $1-\frac{1}{4}+\frac{1}{16}-\mathrm{...}+{\left(-1\right)}^n\frac{1}{4^n}+\mathrm{...}$

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:

  $1-\frac{1}{4}+\frac{1}{16}-\mathrm{...}+{\left(-1\right)}^n\frac{1}{4^n}+\mathrm{...}$

First rewrite the series as Maclaurin series form.

  ${\left(-1\right)}^0\frac{1}{4^0}+{\left(-1\right)}^1\frac{1}{4^1}+{\left(-1\right)}^2\frac{1}{4^2}-\mathrm{...}+{\left(-1\right)}^n\frac{1}{4^n}+\mathrm{...}$

So, the summation form of the series is ${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{1}{4^n}}$ .

Now compare the summation with Maclaurin series summation and yields,

  ${\displaystyle \sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}{\left(x\right)}^n}={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n{\left(\frac{1}{4}\right)}^n}$

So, the point is at $x=\frac{1}{4}$ .

The Maclaurin series of $\frac{1}{1+x}$ is same.

So, the function is $f\left(x\right)=\frac{1}{1+x}$ .

Substitute $x=\frac{1}{4}$ into the function and the sum of the series is $\frac{4}{5}$ .