I am trying to get the coefficients of the power series of $\sqrt{\frac{1+z}{1-z}}$.
Rewriting gets us to $(1 + \frac{2z}{1-z})^{0.5}$. Now using the binomial theorem gets us $\sum_{k=0}^n {\frac{1}{2}\choose k} (\frac{2z}{1-z})^k $ but didnt get far with that.
On
In reply to your comment on where you are blocked, consider that $$ \begin{array}{l} \left( {\sum\limits_{0 \le n} {z^n } } \right)^k = \left( {\frac{1}{{1 - z}}} \right)^k = \left( {1 - z} \right)^{ - k} = \\ = \sum\limits_{0 \le j} {\left( { - 1} \right)^j \left( \begin{array}{c} - k \\ j \\ \end{array} \right)z^j } = \sum\limits_{0 \le j} {\left( \begin{array}{c} j + k - 1 \\ j \\ \end{array} \right)z^j } \\ \end{array} $$
You could write $$\sqrt{\frac{1+x}{1-x}}=\frac{1+x}{\sqrt{1-x^2}}$$ and then expand $$(1+x)(1-x^2)^{-\frac12}$$ and get $$1+x+\frac12x^2+\frac12x^3+\frac38x^4+\frac38x^5+\frac{5}{16}x^6+\frac{5}{16}x^7+...$$
This is convenient since all the coefficients occur in identical pairs...