Chapter 10.2, Problem 11E

To determine

To find: First three nonzero term, general term and interval of convergence of the given function.

Answer to Problem 11E

The first three nonzero terms are, $T_0=x,T_1=x^4\text{and}{T}_2=x^7$

The general term is, $T_n=x^{3n+1}$

And the interval of convergence is $(-1,1)$

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)=\frac{x}{1-x^3}$

Calculation:

As,

  $f\left(x\right)=\frac{x}{1-x^3}=x\left(\frac{1}{1-x^3}\right)$

The Maclaurin series of $\frac{1}{1-x}$ is

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

So, Maclaurin series of $\frac{1}{1-x^3}$ will be

  $\frac{1}{1-x^3}=1+x^3+x^6+\mathrm{....}+{\left(x^3\right)}^n+\mathrm{....}$

So, Maclaurin series of $f\left(x\right)=\frac{x}{1-x^3}$ will be

  $\begin{array}{l}\frac{x}{1-x^3}=x\left(1+x^3+x^6+\mathrm{....}+{\left(x^3\right)}^n+\mathrm{....}\right)\\ =x+x^4+x^7+\mathrm{....}+x^{3n+1}+\mathrm{..}\end{array}$

Hence, the first three term are,

  $T_0=x,T_1=x^4\text{and}{T}_2=x^7$

And the general term of the series is

  $T_n=x^{3n+1}$

Now, for determining the interval of convergence use ration test. So,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\left|\frac{T_n}{T_{n-1}}\right|=\underset{n\to \infty }{\lim }\left|\frac{x^{3n+1}}{x^{3n-2}}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{1}{x^3}\right|\\ =\left|\frac{1}{x^3}\right|\end{array}$

So, the series will converge if $\left|\frac{1}{x^3}\right|<1$

Now checking the convergence at the end points.

Put $x=-1$ in the original series.

  ${\displaystyle \sum _{n=0}^{\infty }{1}^{3n+1}}={\displaystyle \sum _{n=0}^{\infty }1}$

It is a series of infinite sum so it diverges.

Put $x=1$ in the original series.

  ${\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^{3n+1}}$

This series also diverges since it never approaches to zero

So, the series converges if $\left|x^3\right|<1$

Hence, the interval of convergence is $(-1,1)$