Asymptotic expression to $e^{-k^2/2n}$ 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?
 A: 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}
A: $$y=\Bigl(1-{k\over n}\Bigr)^{-(1+n-k)/2}\Bigl(1+{k\over n}\Bigr)^{-(1+n+k)/2}$$
$$\log(y)=-\frac{1+n-k}2 \log\Bigl(1-{k\over n}\Bigr)-\frac{1+n+k}2 \log\Bigl(1+{k\over n}\Bigr)$$
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 $$y \sim e^{-\frac{k^2}{2n}}$$
A: 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
$$
\lim_{n\to\infty}(1+a_n)^{b_n}=e^c\tag1
$$

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 $\sim e^{-\frac{k^2}{2n}}$.
