Chapter 10, Problem 62RE

a.

To determine

Maclaurin series generated by $f\left(x\right)=x^2/\left(1+x\right)$

Answer to Problem 62RE

The Maclaurin series formed by function is,

  $\frac{x^2}{1+x}=x^2-x^3+x^4-\mathrm{...}+{\left(-1\right)}^n{x}^{n+2}+\mathrm{...}$

Explanation of Solution

Given:

The given function is, $f\left(x\right)=x^2/\left(1+x\right)$

Given function,

  $f\left(x\right)=\frac{x^2}{1+x}$

As, the Maclaurin series generated by function $\frac{1}{1+x}$ is,

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

On multiplying above series by $x^2$ results

  $\begin{array}{l}\frac{x^2}{1+x}=x^2(1-x+x^2-\mathrm{...}+{\left(-1\right)}^n{x}^n+\mathrm{...})\\ \Rightarrow \frac{x^2}{1+x}=x^2-x^3+x^4-\mathrm{...}+{\left(-1\right)}^n{x}^{n+2}+\mathrm{...}\end{array}$

Thus, Maclaurin series formed by function is,

  $\frac{x^2}{1+x}=x^2-x^3+x^4-\mathrm{...}+{\left(-1\right)}^n{x}^{n+2}+\mathrm{...}$

b.

To determine

The Series $\frac{x^2}{1+x}=x^2-x^3+x^4-\mathrm{...}+{\left(-1\right)}^n{x}^{n+2}+\mathrm{...}$ converge or diverge at x = 1

Answer to Problem 62RE

At $x=1$ the series diverges.

Explanation of Solution

Given:

The given function is, $f\left(x\right)=x^2/\left(1+x\right)$

From previous subpart, Maclaurin series formed by function $f\left(x\right)=x^2/\left(1+x\right)$ is,

  $\frac{x^2}{1+x}=x^2-x^3+x^4-\mathrm{...}+{\left(-1\right)}^n{x}^{n+2}+\mathrm{...}$

So, the general formula at $x=1$ is $S={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n}$

Now, this series form the sequence of partial sums 0, 1, 0, 1, ….

So, this series diverges.