Asymptotic Difference of Binomial Expansions

Question

Given $x\rightarrow\infty$, I know that $(1-\frac{2\ln2}{x})^{x-2}-(1-\frac{\ln2}{x})^{2(x-1)}$ goes to zero because \begin{align*} \Big(1-\frac{2\ln2}{x}\Big)^{x-2}-\Big(1-\frac{\ln2}{x}\Big)^{2(x-1)} &\sim\exp\Big(-\frac{2\ln2}{x}\cdot x\Big)-\exp\Big(-\frac{\ln2}{x}\cdot 2x\Big) \\ &=0, \end{align*} where I used $\sim$ to mean asymptotic equivalence. How can I find the specific scaling (i.e., the rate of convergence) of the difference?

From the plots I made, it seems to me that it should be something like $O(\frac{1}{x})$, which can be further verified with Wolfram alpha's Big-O domination calculator -- the result is \begin{align*} \Big(1-\frac{2\ln2}{x}\Big)^{x-2}-\Big(1-\frac{\ln2}{x}\Big)^{2(x-1)} &=\frac{\ln4-\ln^22}{4x}-O\Big(\frac{1}{x^2}\Big) \\ &=O\Big(\frac{1}{x}\Big). \end{align*} I tried using Binomial expansions and Taylor expansions but had trouble simplifying. Is there a simpler way to do this (without having to resort to Laurent series which Wolfram claims)?

Answer 1

Let $$A=\left(1-\frac{2 \log (2)}{x}\right)^{x-2}\implies \log(A)=(x-2)\log\left(1-\frac{2 \log (2)}{x}\right)$$

So, by Taylor $$\log(A)=-2\log (2)-\frac{2 ((\log (2)-2) \log (2))}{x}-\frac{4 \left(\log ^2(2) (2\log (2)-3)\right)}{3 x^2}+O\left(\frac{1}{x^3}\right)$$ $$A=e^{\log(A)}=\frac{1}{4}-\frac{(\log (2)-2) \log (2)}{2 x}+\frac{\log ^2(2) (18+\log (2) (3\log (2)-16))}{6 x^2}+O\left(\frac{1}{x^3}\right)$$

Doing the same with

$$B=\left(1-\frac{\log (2)}{x}\right)^{2 (x-1)}$$ $$B=\frac{1}{4}+\frac{2\log (2)-\log ^2(2)}{4 x}+\frac{\log ^2(2) (18+\log (2) (3\log (2)-16))}{24 x^2}+O\left(\frac{1}{x^3}\right)$$

$$A-B=\frac{(2-\log (2)) \log (2)}{4 x}+\frac{\log ^2(2) \left(18+3 \log ^2(2)-16 \log (2)\right)}{8 x^2}+O\left(\frac{1}{x^3}\right)$$

Using your pocket calculator with $x=100$, the exact result is $0.00231558$ while the truncated expression gives $0.00231476$.

Answer 2

An alternative (to the binomial series) is to use $(1-a/x)^{bx-c}=e^{(bx-c)\log(1-a/x)}$, i.e. take the power series for $\log(1-z)$ at $z=a/x$, multiply it by $(bx-c)$, and exponentiate the result (as a power series in $1/x$). Only the last step is nontrivial, especially if you need more than just the main term of the resulting asymptotics, but it can be done mechanically (if you don't need the full expansion in closed form).