Chapter 10.2, Problem 6E

To determine

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

Answer to Problem 6E

The first three nonzero terms are, $T_0=2x,T_1=-\frac{4x^3}{4}\text{and}{T}_2=\frac{4x^5}{15}$

The general term is, $T_n={\left(-1\right)}^n\frac{{\left(2x\right)}^{2n+1}}{\left(2n+1\right)!}$

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

Explanation of Solution

Given:

The given function is,

  $f\left(x\right)=\ln \left(1-x\right)$

Calculation:

As Maclaurin series of $\ln \left(1-x\right)$ is

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

So, Maclaurin series of $f\left(x\right)=\ln \left(1-x\right)$ will be

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

So the first three term are,

  $T_0=-x,T_1=-\frac{x^2}{2}\text{and}{T}_2=-\frac{x^3}{3}$

And the general term of the series is

  $T_n=-\frac{x^n}{n}$

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

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

So, the series will converge if $\left|x\right|<1\text{i}\text{.e}\text{. }-1<x<1$

Now checking the convergence at the end points.

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

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

It is alternating harmonic series so it converges.

Put $x=1$ in the original series.

  ${\displaystyle \sum _{n=1}^{\infty }-\frac{1}{n}}=-{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$

It is harmonic series so it diverges.

So, the series converges when $-1\le x<1$

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