Hi could anyone help me solve this problem.(without expanding the function). I cant seem to get it.
Express the function as a power series
$$\ln\left(\frac{1+x}{1-x}\right)$$
I managed to split the $\ln(1+x)-\ln(1-x)$ apart but i do not know how to proceed from here.
Based on $$ln(1+x)=\sum _{ n=1 }^{ \infty }{ { \left( -1 \right) }^{ n-1 } } \frac { { x }^{ n } }{ n } ,\quad \left| x \right| <1$$ we get $$ln(1-x)=-\sum _{ n=1 }^{ \infty }{ \frac { { x }^{ n } }{ n } } ,\quad \left| x \right| <1$$
Therefore,
$$ln\frac { 1+x }{ 1-x } =2\sum _{ k=0 }^{ \infty }{ \frac { x^{ 2k+1 } }{ 2k+1 } } ,\quad \left| x \right| <1$$