Maclaurin Series Expansion nth term expression

Question

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.

Answer 1

With $$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots$$ $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots$$ gives $$\ln(1-x)-\ln(1+x)=-2x-\frac{2x^3}{3}-\frac{2x^5}{5}-\cdots$$ or $$\ln\dfrac{1-x}{1+x}=\sum_{n=0}^\infty\dfrac{-2x^{2n+1}}{2n+1}$$ valid for $|x|<1$.

Answer 2

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 $$\ln\frac{x+1}{x-1}= \ln (1+x)-\ln (x-1)$$

$$=x-\frac{x^2}{2}+\frac{x^3}{3}...+x+\frac{x^2}{2}+\frac{x^3}{3}+....$$

$$= 2x+(2/3) x^3 + 2/5 (x^5) +(2/7)x^7 +..... $$

$$= \sum_{n=1}^{\infty} (\frac{2}{2n-1})x^{2n-1}$$