Below question is 34 on page 149 of Edexcel Pearson's FP2 (Further Pure Mathematics 2.
Show that for $x>1$, $$\ln(x^2-x+1)+\ln(x+1)-3\ln(x)=\frac{1}{x^3}-\frac{1}{2x^6}+...+\frac{(-1)^{n-1}}{nx^{3n}}$$
I have let $g(x)$ equal to the initial function, then done derivatives of $g$ to try to go towards a Taylor expansion around point $x=1$.
I then have $g(1)=\ln(2)$, $g'(1)=-\frac{3}{2}$, $g''(1)=\frac{11}{4}$.
However my coefficients seem to be wrong.
I also see that $g(0)$ is not defined.
I am seeking:
- An answer to the question using a Maclaurin or Taylor expansion with full working.
- An explanation of which set of values of $x$ can be used for a Taylor expansion.
If you are allowed to use well known series, you don't have to go through the trouble of setting up the series manually (via derivatives).
Simplify by using properties of logarithms: $$\begin{align}\ln\left(x^2-x+1\right)+\ln\left(x+1\right)-3\ln(x) & = \ln\frac{\left(x^2-x+1\right)\left(x+1\right)}{x^3} \\[4pt] & = \ln\frac{x^3+1}{x^3} \\[5pt] & = \ln\left(1+\frac{1}{x^3}\right)\end{align}$$ And recall the standard series, for $|a|<1$: $$\ln(1+a) = a - \frac{a^2}{2}+\frac{a^3}{3}-\ldots+(-1)^{n+1}\frac{a^n}{n}+\ldots$$ Note that $x>1 \implies 0<x^{-3}<1$; now use $a= x^{-3}$.