I need to prove that $$1-\frac{(A-1)^2}{2A}\ln \frac{A+1}{A-1}$$ approximately equals $\dfrac{2}{A+2/3}$.
I think that we can expand the $\ln$ to $2(1/A+1/(3A^3)+\dots)$ and so the first term multiplied by the polynomial and all reduces to $2/A$ since $1/A^2$ -> really quickly. Can anyone help me reduce this equation?
It helps to take $x=1/A$. Then the function becomes $$1-\frac{x(1/x-1)^2}{2}\ln \frac{1+x}{1-x}$$ The Taylor expansion at $x=0$ (best found with CAS) is $$2x-\frac43 x^2 + \frac23 x^3 + \dots $$ which agrees up to second order with $$\frac{2}{1/x+2/3} = 2x-\frac43 x^2 + \frac89 x^3 -\dots$$