How to find Laurent series?

Question

I have a function $f(z)={z^2}/\sqrt{1+z^2}$. I need to expand it neighborhood of $z=\infty$ Here i made used $z^2+1=u^2$ then i had $f(u)=u-1/u$ but these substition seems useless to me. However, i cannot find any way to expand this function. Any suggestions?

Answer 1

I expanded at $x=0$ the function $$\frac{1}{ \sqrt{x^2+1}}=\sum _{n=0}^{\infty } \frac{x^{2 n} \left((-1)^n \Gamma \left(n+\frac{1}{2}\right)\right)}{\sqrt{\pi } n!}$$ Then I divided both sides by $x$ $$\frac{1}{ x\sqrt{x^2+1}}=\sum _{n=0}^{\infty } \frac{x^{2 n-1} \left((-1)^n \Gamma \left(n+\frac{1}{2}\right)\right)}{\sqrt{\pi } n!}$$ And finally substituted $x=\frac{1}{z}$ $$\frac{z^2}{\sqrt{z^2+1}}=\sum _{n=0}^{\infty } \frac{z^{1-2 n} \left((-1)^n \Gamma \left(n+\frac{1}{2}\right)\right)}{\sqrt{\pi } n!}$$ Hope this helps