<s>[INST] Fraction/Binomial Coefficient Estimate Let $k \leq k+x <n.$ Prove
$$(1)\ \ \ \ \ \ \ \ \ \ \ \ \left( \frac{n-k-x}{n-x}\right)^x \leq \frac{{n-x \choose k}}{{n \choose k}} \leq \left(\frac{n-k}{n}\right)^x.$$
(This is part of formula 1.23 in Stasys Jukna's extremal combinatorics.)
What I have tried:
(A) I tried to apply Stirling's formula to the factorials. This version of Stirling states
$$\Gamma(n+1) = \sqrt{\tau}\frac{n^{n+1/2}}{e^n} e^{\varepsilon_n},$$
with some kind of error bounds on $\varepsilon_n,$ e.g. $0 < \varepsilon_n < \frac{1}{12n}$ if $n \geq 1.$ Applying Stirling's formula to the first inequality in (1), we get an equivalent inequality
$$\text{ERROR}\cdot \left(\frac{n-k}{n-k-x}\right)^{n-k+1/2} \leq \left(\frac{n}{n-x}\right)^{n+1/2},$$
where the ERROR term is from Stirling's approximation. I don't know what to do from here.
(B) I tried expanding and applying a logarithm. Using the power series for $\ln(1+x),$ I found that (1) is equivalent to
$$\frac{k}{n}x + \left(\frac{k}{n}\right)^2 x + ... \leq \left(\frac{1}{n} + \frac{1}{n-1} + ... + \frac{1}{n-k+1} \right)x + \left(\frac{1}{n^2} + ... + \frac{1}{(n-k+1)^2} \right)x^2 + ... \leq \frac{k}{n-x}x + \left(\frac{k}{n-x}\right)^2 x + ....$$
I don't know what to do from here either.
 [/INST] It may be helpful to adjust the $\dfrac{{n-x \choose k}}{{n \choose k}}$ expression so each of the three items is the product of $x$ terms:

*

*$\dfrac{{n-x \choose k}}{{n \choose k}}= \frac{n-x}{n}\frac{n-x-1}{n-1}\frac{n-x-2}{n-2}\cdots \frac{n-x-k+1}{n-k+1} = \frac{n-k}{n}\frac{n-k-1}{n-1}\frac{n-k-2}{n-2}\cdots \frac{n-k-x+1}{n-x+1} = \dfrac{{n-k \choose x}}{{n \choose x}}$ with the second product having $x$ terms


*$\frac{n-k-x}{n-x} = 1- \frac{k}{n-x}\le  \frac{n-k-i}{n-i}  = 1- \frac{k}{n-i}\le  \frac{n-k}{n}  = 1- \frac{k}{n}$ when $0 \le i \le x \le n-k$


*so $\left(\frac{n-k-x}{n-x}\right)^x \le  \frac{n-k}{n}\frac{n-k-1}{n-1}\frac{n-k-2}{n-2}\cdots \frac{n-k-x+1}{n-x+1} \le \left(\frac{n-k}{n}\right)^x$
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