Series Expansion of $\sqrt{\frac{1}{x}-1}$

Question

I am trying to expand this function around $x_0=0$.

Using Mathematica I get $$\sqrt{\frac{1}{x}-1}=\frac{1}{\sqrt{x}}-\frac{\sqrt{x}}{2}-\frac{x^{\frac{3}{2}}}{8}+\mathcal{O}(x^{\frac{5}{2}})$$

I somehow manage to calculate the first term by hand, even though I don't know if it is correct.

But I don't understand how to calculate the higher order terms.

Any help would be appreciated.

Answer 1

Try

$$ \sqrt{\frac{1}{x}-1} = \frac{1}{\sqrt{x}}\sqrt{1-x} $$

and

$$ \sqrt{1-x} = 1+\frac{1}{2}(-x)+\frac{1}{2}\left(\frac{1}{2}-1\right)\frac{(-x)^2}{2!}+\cdots $$

Answer 2

$$\sqrt{{1\over x}-1}=(1-x)^{1/2}x^{-1/2}$$

$$\begin{align} (1-x)^{1/2}&=\sum_{k=0}^{\infty}\binom{1/2}{k}(-1)^kx^k\\ &={1}-{\frac12\over1!}x^1+{\frac12(\frac12-1)\over2!}x^2-{\frac12(\frac12-1)(\frac12-2)\over3!}x^3\dots \end{align}$$

Just multiply this series by $x^{-1/2}$ and you'll get the answer