How would i find a series representation for ln(x) from the series of ln(1+x)? which is: [(-1)^n * x^n] / n
$\displaystyle \ln(x+1)=\int\frac{dx}{x+1}=\int\sum_{n=0}^{\infty }(-1)^nx^n=\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{n+1}$
Now to find ln(x) in terms of this
Kalgotta was correct the first step is to see that $\displaystyle \ln(x)=\ln(1+(x-1))$
So by letting $\displaystyle \lambda=x-1$ we get
$\displaystyle \ln(1+\lambda)=\sum_{n=0}^{\infty}\frac{(-1)^n\lambda^{n+1}}{n+1}$
Substituting back in what u equals we get
$\displaystyle \ln(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}$
To verify we see that
$\displaystyle \ln(x)=\int\frac{dx}{x}=\int\sum_{n=0}^{\infty}(-1)^n(x-1)^ndx=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}$
QED