My task is to show that the Maclaurin series of $\ln\frac{x+1}{x-1}$ is given as $\sum_{n=1}^{\infty} (\frac{2}{2n-1})x^{2n-1}$
My approach is using the standard expansion of $\ln(x+1)$ to get $x-\frac{x^2}{2}+\frac{x^3}{3}$
I got $$\ln(x-1) = -x-\frac{x^2}{2}-\frac{x^3}{3}$$
After this point I'm a little confused as to how to proceed. My guess would be to subtract $\ln(x-1)$ from $\ln(x+1)$ but I still don't understand how to get this form or what exactly it's asking.
Any help would be greatly appreciated.
Best Answer
You have $$ \ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+...$$
and $$\ln (x-1) = -x-\frac{x^2}{2}-\frac{x^3}{3}+...$$
Thus $$\ln\frac{x+1}{x-1}= \ln (1+x)-\ln (x-1)$$
$$=x-\frac{x^2}{2}+\frac{x^3}{3}...+x+\frac{x^2}{2}+\frac{x^3}{3}+....$$
$$= 2x+(2/3) x^3 + 2/5 (x^5) +(2/7)x^7 +..... $$
$$= \sum_{n=1}^{\infty} (\frac{2}{2n-1})x^{2n-1}$$