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?

One can write $1+y=1/t^2$ so that $t\to 0^+$ and then the given expression, say $F(y) $, transforms into $$F(y) =\left(\frac{1}{t^2}-\frac{1}{3}\right)\log\frac{1+t}{1-t}-\frac{2}{t}\tag{1}$$ and we need to find the limit of $y^{3/2}F(y)$ or $$\frac{(1-t^2)^{3/2}}{t^3}F(y)\tag{2}$$ Since $t\to 0^+$ the desired limit is the same as that of $F(y) /t^3$.
Using Taylor series we can write $F(y) $ as $$2\left(\frac{1}{t^2}-\frac{1}{3}\right)\left(t+\frac{t^3}{3}+\frac{t^5}{5}+o(t^5)\right)-\frac{2}{t}$$ And this can be written as $$2\left(\frac{1}{t}+\frac{t}{3}+\frac{t^3}{5}-\frac{t}{3}-\frac{t^3}{9}+o(t^3)-\frac{1}{t}\right)=\frac{8t^3}{45}+o(t^3) $$ It follows that $F(y) /t^3$ tends to $8/45$.
One can observe that the substitution $1+y=1/t^2$ helps here a lot as it simplifies the log term to a well known Taylor series.
