Asymptotic expression to $e^{-k^2/2n}$

Question

I want to show that $$\Bigl(1-{k\over n}\Bigr)^{-(1+n-k)/2}\Bigl(1+{k\over n}\Bigr)^{-(1+n+k)/2}$$ is asymptotic to $e^{-k^2/2n}$.

I, in my earlier post, was satisfied with the given answer. But only now I noticed that the answer provided has a flaw since though it claims to produce $e^{-k^2/2n}$, it actually produces $e^{-k^2/n}$.

Any thoughts?

Answer 1

You have \begin{align} \Bigl(1-{\tfrac kn}\Bigr)^{-\frac{1+n-k}2}\Bigl(1+{\tfrac kn}\Bigr)^{-\frac{1+n+k}2} &= \exp\left[-\tfrac{1+n-k}2\,\log\left(1-\tfrac kn\right)-\tfrac{1+n+k}2\,\log\left(1+\tfrac kn\right)\right]\\ \ \\ &=\exp\left[-\tfrac{1+n-k}2\,\left(-\tfrac kn-\tfrac{k^2}{2n^2}+o(\tfrac{k^3}{n^3})\right)-\tfrac{1+n+k}2\, \left(\tfrac kn-\tfrac{k^2}{2n^2}+o(\tfrac{k^3}{n^3})\right)\right]\\ \ \\ &=\exp\left[-\tfrac{1+n-k}2\,\left(-\tfrac kn-\tfrac{k^2}{2n^2}+o(\tfrac{k^3}{n^3})\right)-\tfrac{1+n+k}2\, \left(\tfrac kn-\tfrac{k^2}{2n^2}+o(\tfrac{k^3}{n^3})\right)\right]\\ \ \\ &=\exp\left[-\tfrac{k^2}n+(1+n)\tfrac{k^2}{2n^2}+o(\tfrac1{n^2}) \right]\\ \ \\ &=\exp\left[-\tfrac{k^2}{2n}+\tfrac{k^2}{2n^2}+o(\tfrac1{n^2}) \right]\\ \ \\ &=\exp\left[-\tfrac{k^2}{2n}+o(\tfrac1{n^2}) \right]\\ \ \\ \end{align}

Answer 2

$$y=\Bigl(1-{k\over n}\Bigr)^{-(1+n-k)/2}\Bigl(1+{k\over n}\Bigr)^{-(1+n+k)/2}$$ $$\log(y)=-\frac{1+n-k}2 \log\Bigl(1-{k\over n}\Bigr)-\frac{1+n+k}2 \log\Bigl(1+{k\over n}\Bigr)$$

Use the Taylor expansions of $\log\Bigl(1\pm{k\over n}\Bigr)$; replace, expand and simplify. You should arrive at $$\log(y)=-\frac{k^2}{2n}+\frac{k^2}{2n^2}+\cdots$$ making $$y \sim e^{-\frac{k^2}{2n}}$$

Answer 3

Letting $f=1+a_n$ and $g=b_n$ in this answer, we get the following

Lemma: Suppose that $\lim\limits_{n\to\infty}a_n=0$ and $\lim\limits_{n\to\infty}a_nb_n=c$, then $$ \lim_{n\to\infty}(1+a_n)^{b_n}=e^c\tag1 $$

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If $\lim\limits_{n\to\infty}\frac{k^2}{2n}=\alpha$, then $$ \begin{align} \lim_{n\to\infty}\left(1-\frac kn\right)^{-\frac{n-k+1}2}\left(1+\frac kn\right)^{-\frac{n+k+1}2} &=\lim_{n\to\infty}\left(1-\frac{k^2}{n^2}\right)^{-\frac{n+1}2}\,\frac{\lim\limits_{n\to\infty}\left(1-\frac kn\right)^{\frac k2}}{\lim\limits_{n\to\infty}\left(1+\frac kn\right)^{\frac k2}}\\ &=e^{\alpha}\,\frac{e^{-\alpha}}{e^\alpha}\\[9pt] &=e^{-\alpha}\tag2 \end{align} $$ Which is $\sim e^{-\frac{k^2}{2n}}$.