We know the following:
$$\sqrt{x+e} = \sqrt{x}(1+\frac{e}{2x}-\frac{e^2}{8x^2}+\frac{e^3}{16x^3}-\cdots)$$ as $\frac{e}{x}\rightarrow 0$. I am wondering whether this series is uniformly convergent for different $x$ when $e$ goes to zero?
I am tring to prove this based on the radius of convergence but have no idea to do it. Could anyone give me a hint? thanks!
It looks like you factored out $\sqrt{x}$ from $\sqrt{x+e}$ and then applied the binomial theorem to get the $1/2$ power of the other factor $1+\frac{e}{x}.$
With $u=e/x,$ the binomial theorem gives a series with radius of convergence $1$ as may be seen via the ratio test on the coefficients. [Note the fourth power term is $-5/128$ which means the pattern of coefficients is not just reciprocal powers of $2$ even though that appears so from up to your cubic term.]
Anyway in terms of $x,e$ with $e\to 0$ it seems for fixed $x$ you get absolute convergence provided $|e/x|<1.$