How to prove that $\frac{k}{1-\beta^2}-\frac{k}{\sqrt{1-\beta^2}}=\frac{k\beta^2}{2}+O(\beta^4)$ for small $\beta$?

Question

I need to prove that

$$\frac{k}{1-\beta^2}-\frac{k}{\sqrt{1-\beta^2}}=\frac{k\beta^2}{2}+O(\beta^4)$$

for small $\beta$.

I have tried out the Maclaurin expansion of both expressions on the LHS, but I'm not getting the RHS result from that. The expansion of the second term in LHS in particular is quite complicated and I think there must be a simpler way to prove. Would like some help on this!

Answer 1

It may be simpler to avoid expanding $\frac1{\sqrt{1-\beta^2}}$. Note that $\frac{1}{1-\beta^2}=1+O(\beta^2)$ and,

$$\sqrt{1-\beta^2}=1-\frac12{\beta^2}+O(\beta^4)\implies 1-\sqrt{1-\beta^2}=\frac12{\beta^2}+O(\beta^4)$$

Thus,

$$\frac{k}{1-\beta^2}-\frac{k}{\sqrt{1-\beta^2}} =\frac{k}{1-\beta^2}\left( 1- \sqrt{1-\beta^2}\right)$$ $$=k(1+O(\beta^2))(\frac12{\beta^2}+O(\beta^4))=\frac{k\beta^2}{2}+O(\beta^4)$$

Answer 2

$$\frac1{1-t}=1+t+o(t)$$

$$\frac1{\sqrt{1-t}}=1+\frac t2+o(t)$$

and by subtraction,

$$\frac t2+o(t).$$