Limit of $\left(y+\frac23\right)\mathrm{ln}\left(\frac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}$ We consider the limit for large $y$ of the following expression :
$$\left(y+\frac23\right)\mathrm{ln}\left(\dfrac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}.$$
Many references state that the large $y$ behaviour of this last expression is $y^{-3/2}$. I have been trying to show this in any way possible but I'm only hitting dead ends. Wolframalpha gives a Puiseux series of that expression which makes this behaviour explicit, but I do not understand how to get to this series either. Could you please help me on this?

Considering $$A=\left(y+\frac23\right)\log\left(\dfrac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}$$ first simplify
$$A=\left(y+\frac{2}{3}\right) \log \left(\frac{y+2 \sqrt{y+1}+2}{y}\right)-2
   \sqrt{y+1}$$ For large values of $y$, you could start using $y=\frac 1x$ to make
$$A=\frac{(2 x+3) \log \left(2 x+2 \sqrt{x (x+1)}+1\right)-6 \sqrt{x (x+1)}}{3 x}$$ Now, Taylor series
$$\sqrt{x (x+1)}=x^{1/2}+\frac{x^{3/2}}{2}-\frac{x^{5/2}}{8}+\frac{x^{7/2}}{16}+O\left(x^{9/2}\right)$$
$$ \log \left(2 x+2 \sqrt{x (x+1)}+1\right)=2 x^{1/2}-\frac{x^{3/2}}{3}+\frac{3 x^{5/2}}{20}-\frac{5 x^{7/2}}{56}+O\left(x^{9/2}\right)$$
$$A=\frac{8 x^{3/2}}{45}-\frac{4 x^{5/2}}{35}+O\left(x^{7/2}\right)$$ Back to $y=\frac 1x$ gives the result.
As I used to say : we are always closer to $0$ than to $\infty$
