$$\frac{1}{r}\left(1+\frac{2\epsilon \cos\theta}{r}\right)^{-1/2}$$ I was told by using Taylor expansion I could get
$$1-\frac{2\epsilon \cos\theta}{r}$$ with term of order $\epsilon^2$. Can someone explain to me how?


That's wrong.
$$ \dfrac{1}{r} \left( 1 + \dfrac{2\epsilon \cos(\theta)}{r}\right)^{-1/2} = \dfrac{1}{r} - \dfrac{\cos(\theta)}{r^2} \epsilon + O(\epsilon^2)$$ That comes from the binomial series: $$ (1 + x)^p = 1 + p x + O(x^2)$$


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