Question

In Problems 11-16 use an appropriate series in (2) to find the Maclaurin series of the given function. Write your answer in summation notation.

$\mathit{l}\mathit{n}\mathbf{(}\mathbf{1}\mathbf{-}\mathit{x}\mathbf{)}$

Short answer

$\ln (1-x)=\sum _{n=1}^{\infty }\frac{-1}{n}{x}^n$

Step-by-step solution

Definition

Maclaurin seriesare a type of series expansion in which all terms are nonnegative integer powers of the variable.

Find Maclaurin series

Consider$\ln (1-x) ----\left(1\right)$

Consider the Maclaurin series of$\ln (1+x)$ given as below:

$$\begin{gathered} \ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots \\ =\sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}}{n}{x}^n\dots \dots \left(2\right) \end{gathered}$$

Replace$x$ by$-x$ in equation (2) to equation (1) in the following form:

$$\begin{gathered} \ln (1-x)=\ln (1+(-x\left)\right) \\ =(-x)-\frac{{(-x)}^2}{2}+\frac{{(-x)}^3}{3}-\frac{{(-x)}^4}{4}+\dots \\ =-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+\dots \\ =\sum _{n=1}^{\infty }\frac{-x^n}{n} \end{gathered}$$

Thus, the Maclaurin series for function $\ln (1-x)=\sum _{n=1}^{\infty }\frac{-1}{n}{x}^n$.