Chapter 9.2, Problem 6E

To determine

To find: To find the general term of the Maclaurin series of the given function.

Answer to Problem 6E

The McLaurin series is $T_n=-\frac{x^n}{n}$ and series always converges for $(-1,1)$ .

Explanation of Solution

Given:

The given series $f(x)=\ln (1-x)$ .

Formula used:

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

Calculation:

Given the given function is $f(x)=\ln (1-x)$ .

By replacing $x=-x$ ,

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

The first three non- zero terms of the series are $T_1=-x,T_2=-\frac{x^2}{2},T_3=-\frac{x^3}{3}$ .

The general term of the series is given by $T_n=-\frac{x^n}{n}$ .

The interval of convergence uses the ratio test.

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

So the ratio tells us that if $L<1$ ,the series will converge and if $L>1$ ,the series will diverge. We don’t know that what will happen so we have $\left|x\right|<1$ , $-1<x<1$ . The way to determine the convergence at end points is to simply plug them into the power series and see if the convergence or divergence using any test necessary.

The series converges between $-1\le x<1$ .