show $(1+\frac{a}{n})^{n-k}=e^a(1-\frac{a(a+k)}{2n})+o(\frac{1}{n})$ for a fixed nonnegative integer $k$ as $n\to\infty$?

Question

How to show $(1+\frac{a}{n})^{n-k}=e^a(1-\frac{a(a+k)}{2n})+o(\frac{1}{n})$ for a fixed nonnegative integer $k$ as $n\to\infty$?

We already known that $(1+\frac{a}{n})^{n}\to e^a$ as $n\to\infty$, but I don't know how to deal with $(1+\frac{a}{n})^{-k}$. Could someone kindly help? Thanks.

http://sites.stat.psu.edu/~dhunter/asymp/fall2004/lectures/edgeworth.pdf

Answer 1

$$A=\left(1+\frac{a}{n}\right)^{n-k}\implies \log(A)=(n-k) \log\left(1+\frac{a}{n}\right)$$ Now, Taylor series for large $n$ $$\log\left(1+\frac{a}{n}\right)=\frac{a}{n}-\frac{a^2}{2 n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(A)=a-\frac{a (a+2 k)}{2 n}+O\left(\frac{1}{n^2}\right)=a\left(1 -\frac{ (a+2 k)}{2 n}\right)+O\left(\frac{1}{n^2}\right)$$ Now, using Taylor again $$A=e^{\log(A)}=e^a-\frac{a e^a (a+2 k)}{2 n}+O\left(\frac{1}{n^2}\right)=e^a\left(1- \frac{a(a+2 k)}{2 n}\right)+O\left(\frac{1}{n^2}\right)$$

I wonder if there could be a typo in the book ($k$ instead of $2k$)

Edit

Using $a=10$, $k=100$, $n=10000$, the "exact" value is $A\approx 19832.024$; the approximation given above gives $\frac{179 }{200}e^{10}\approx 19713.687$ while what is given in the book gives $ \frac{189 }{200}e^{10}\approx 20815.010$.