Let $p,k$ be natural numbers with $p\ge k$, show that
$$ \frac{(p-k)!}{(p+k)!} \le \frac{1}{p^{2k}} \left(\frac{e}{2} \right)^{2k^2/p}. $$
The text where I come across this says to use Stirling's formula (which I assume is the same as Stirling's inequality, please let me know if this is incorrect.)
Attempt #1. Using Stirling's inequality we have
\begin{align*} \frac{(p-k)!}{(p+k)!} &\le \frac{ \frac{e}{e^{p-k}} (p-k)^{p-k+1/2} } { \frac{\sqrt{2 \pi} }{e^{p+k} } (p+k)^{p+k+1/2} } \end{align*}
But if choose $p=20$ and $k=1$ we get (using mathematica's arbitrary precision arithmetic) the RHS is greater than our goal. So this strategy is doomed to fail. I thought maybe the inequality is off by a constant multiplicative factor but I haven't found a counter example. Any ideas?
Almost six years too late !
$$\text{lhs }=\frac{(p-k)!}{(p+k)!}\qquad \text{and} \qquad \text{rhs } =\frac{1}{p^{2k}} \left(\frac{e}{2} \right)^{2k^2/p}$$ Take logarithms and use Stirling approximation for the lhs and continue with Taylor expansions $$\log(\text{lhs})=-2 k \log (p)-\frac{k}{p}+\frac{2 k^3+k}{6 p^2}-\frac{k^3}{3 p^3}+O\left(\frac{1}{p^4}\right)$$ $$\log(\text{rhs})=-2 k \log (p)+\frac{2 k^2 (1-\log (2))}{p}$$ $$\log(\text{rhs})-\log(\text{lhs})=\frac{k (1+2 k (1-\log (2)))}{p}-\frac{k \left(2 k^2+1\right)}{6 p^2}+O\left(\frac{1}{p^3}\right)$$ Now, using $x=e^{\log(x)}$ $$\frac{\text{rhs} } {\text{lhs} }=1+\frac{k (1+2 k (1-\log (2)))}{p}+O\left(\frac{1}{p^2}\right)$$ So, the inequality holds.