Thread: [SOLVED] McLaurin series for ln(...)

1. [SOLVED] McLaurin series for ln(...)

Need McLaurin series of $\displaystyle \ln{\left(\frac{1+2x}{1-3x+2x^2}\right)}$

Tried to make it like $\displaystyle \ln\left({1+x}\right)$ and found $\displaystyle \ln\left(1+\frac{5x-2x^2}{1-3x+2x^2}\right)$

So, can i put that $\displaystyle \;\left(\frac{5x-2x^2}{1-3x+2x^2}\right)$ into $\displaystyle \sum\limits_{n= 1}^\infty\;\frac{{\left(-1\right)}^{n-1}}{n}x^n$

if i can not, how can i solve that problem?

2. Originally Posted by Lafexlos
Need McLaurin series of $\displaystyle \ln{\left(\frac{1+2x}{1-3x+2x^2}\right)}$

Tried to make it like $\displaystyle \ln\left({1+x}\right)$ and found $\displaystyle \ln\left(1+\frac{5x-2x^2}{1-3x+2x^2}\right)$

So, can i put that $\displaystyle \;\left(\frac{5x-2x^2}{1-3x+2x^2}\right)$ into $\displaystyle \sum\limits_{n= 1}^\infty\;\frac{{\left(-1\right)}^{n-1}}{n}x^n$

if i can not, how can i solve that problem?
I'd start by using the log rule ln(A/B) = ln(A) - ln(B).

3. That looks much better.
Thx for help. =)