Finding power series of function could anyone help me answer question?
$$F(x)=\ln\left(\dfrac{7+x}{7-x}\right)$$ Find a power series representation for the function.
 A: We will assume that $7-x$ and $7+x$ are positive. Note that 
$$\ln\left(\frac{7+x}{7-x}  \right)=\ln\left(\frac{1+x/7}{1-x/7}  \right)=\ln(1+x/7)-\ln(1-x/7).$$
You know the power series expansion of $\ln(1+t)$. Substitute $t=x/7$, $t=-x/7$ and subtract. 
Remark: The trick goes back at least to Euler. Instead of using $7$, use $3/2$, and let $x=1/2$. Note that $\frac{3/2+1/2}{3/2-1/2}=2$. We can use the idea to find a reasonably quickly convergent series for $\ln 2$. 
A: It is very helpful to keep the following series in mind with problems like this:
$$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n$$
Integrating, we find
$$\ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}$$
You can now readily use the fact that
$$\ln\frac{7+x}{7-x}=\ln\left(1+\frac{x}{7}\right)-\ln\left(1-\frac{x}{7}\right)$$
The resulting series will converge for $|x|<7$.
A: When I was young (that is to say a loooong time ago), one of the series the professor asked us to always remember (because of its extreme simplicity) is precisely $$\ln\left(\frac{1+x}{1-x}  \right)=2\sum_{n=0}^{\infty}\frac {x^{2n+1}}{2n+1}$$ Thank you for remembering me my youth
