Understanding $\log p/q = \sum_{k=1}^{\infty} \frac{1}{2k-1}(\frac{p-q}{p+q})^{2k-1}$ Problem

This problem comes straight from the Taylor Formula Chapter of Edwin Wilson's Advanced Calculus Textbook:



The part that I am concerned with is part $(\gamma)$.
What I need:

Either a hint pointing me in the right direction or a guide to a resources that will hellp illustrate why this relationship is true.
Personal Work and Ideas:

The problem comes down to proving that:
$$\log\frac{p}{q} = 2\bigg[\sum\limits_{k=1}^{\infty}\frac{1}{2k-1}(\frac{p-q}{p+q})^{2k-1}\bigg]$$
which on its face seems to be some manipulation of the series
$$\text{ (1)   }\log x = \log a + \sum\limits_{k=1}^\infty \frac{(-1)^{k-1}}{k}  (\frac{x-a}{a})^k$$

Origin of

This formula of course comes from taylor expansion of $\log x$ and is derived from these applying these two pieces of information:
$$f^{(k)}(x) = \begin{cases} 
      \log x & k=0 \\
      \frac{(-1)^{k-1}}{(x)^k}(k-1)!& k>0\\ 
   \end{cases} $$
and
$$f(x) = f(a) + \sum\limits_{k=1}^{\infty} f(a)\frac{(x-a)^k}{k!}$$

Some ideas I have tried are the following (Caution this mostly reads like scratch work.):

IDEA 1

Idea : Set $x=p$ and $a=q$
Motivation:
As a first attempt I am trying to see how far simple substitution takes me. So I am going to plug in $x=p$ and $a=q$ to see how far it goes..
Attempt
Assume that we plug $x=p$ and $a=q$ in to (1) then we obtain:
$$\log (\frac{p}{q}) = \sum\limits_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}(\frac{p-q}{q})^{k}$$
Problems
The biggest problems with this approach are two fold.

*

*No introduction to $p+q$ as a denominator term.

*The k-values in parity with what the solution is giving us (all k-values of the solution are odd).



IDEA 2

Idea : Sum pairs of summands
Motivation:
The issue with the parity of the $k$ values seems to be addressed if we sum pairs of items. This possibly might introduce $p+q$ to the denominator.
Attempt
Let $b_k =\frac{(-1)^{k-1}}{k}(\frac{p-q}{q})^{k} $ then $$b_{2k}+b_{2k+1} = \frac{(-1)^{2k-1}}{2k}(\frac{p-q}{q})^{2k} +\frac{(-1)^{2k}}{2k+1}(\frac{p-q}{q})^{2k+1}   $$
$$=(\frac{p-q}{q})^{2k}\bigg[-\frac{1}{2k} +\frac{\frac{p-q}{q}}{2k+1}\bigg]$$
$$=(\frac{p-q}{q})^{2k}\bigg[\frac{-q(2k+1) +(p-q)(2k)}{(2qk)(2k+1)}\bigg]$$
Problems:
There doesn't seem to be a good way that this turns to the denominator to $p+q$.


IDEA 3

Idea: Take the difference of to taylor series. Namely $\log p$ and $\log q$ with the center being at p+q.
Motivation
The goal of this is to introduce the $p+q$ term to the denominator.
Attempt
$$\log p = \log(p+q) + \sum\limits_{k=1}^{\infty} \frac{(-1)^{k-1}}{k}  (\frac{p-(p+q)}{p+q})^k$$
or
$$\log p = =\log(p+q) + \sum\limits_{k=1}^{\infty} \frac{(-1)^{2k-1}}{k}  (\frac{q}{p+q})^k$$
$$\log q=\log(p+q) + \sum\limits_{k=1}^{\infty} \frac{(-1)^{2k-1}}{k}  (\frac{p}{p+q})^k$$
which after subtraction gives us this:
$$\log \frac{p}{q}= \sum\limits_{k=1}^{\infty} \frac{(-1)^{2k-1}}{k}\bigg[  (\frac{q}{p+q})^k-(\frac{p}{p+q})^k\bigg]= \sum\limits_{k=1}^{\infty} \frac{(-1)^{2k-1}}{k}(\frac{1}{p+q})^k\bigg[  q^k-p^k\bigg]$$
Notes:
Since this is my current idea and I am muddling through, I will rename the problem section notes. I am not sure if this route will work or not. Might end up having a problem with the coeffients. I am still thinking through possible next steps.

 A: Hint: $\displaystyle
\frac pq = \bigg( 1 + \frac{p-q}{p+q} \bigg) \bigg/ \bigg( 1 - \frac{p-q}{p+q} \bigg)
$.
A: The hint is:  Don't get confused by $p$ and $q$, use $x=p/q$ and write
$$
\ln x 
= 2\sum_{k=1}^{\infty}
  \frac{1}{2k-1} \cdot \left(\frac{x-1}{x+1}\right)^{2k-1}
\!\!\!=\; 2 \operatorname{artanh} \frac{x-1}{x+1}
$$
Then you are left with exploring the properties or $\operatorname{artanh}$, the inverse of the hyperbolic tangent
$$
x\mapsto \frac{\sinh x}{\cosh x}
= \frac{e^x-e^{-x}}{e^x+e^{-x}}
= \frac{e^{2x}-1}{e^{2x}+1}
$$
The relation between $\ln$ and $\operatorname{artanh}$ is easily derived from there; over $\Bbb R$ or over $\Bbb C$ depending on your preference.
The Taylor series for $\operatorname{artanh}$ follows from $$\operatorname{artanh}(ix) = i\arctan x$$ and from the Taylor series of $\arctan$ which in turn follows from
$$\arctan x = \int_0^x \frac{dt}{1+t^2}
$$
and from $$\sum_{k=0}^\infty x^k = \frac1{1-x} \qquad\text{for}\quad|x|<1$$
Maybe some complex analysis is overkill, but to me it seems to provide more insight than getting lost in struggling with low-level manipulations.  However I don't know what knowlege the textbool implies; but at least it's about complex analysis.
