Putnam 1996, Problem B-2

Show that for every positive integer \( n \),  
\[
\left( \frac{2n-1}{e} \right)^{\frac{2n-1}{2}} < 1 \cdot 3 \cdot 5 \cdots (2n-1) < \left( \frac{2n+1}{e} \right)^{\frac{2n+1}{2}}.
\]

Solution:
By estimating the area under the graph of \(\ln x\) using upper and lower rectangles of width 2, we get  
\[
\int_1^{2n-1} \ln x \, dx \leq 2(\ln 3 + \cdots + \ln(2n - 1)) \leq \int_3^{2n+1} \ln x \, dx.
\]  
Since \(\int \ln x \, dx = x \ln x - x + C\), we have, upon exponentiating and taking square roots,  
\[
\left( \frac{2n - 1}{e} \right)^{\frac{2n - 1}{2}} < (2n - 1)^{\frac{2n - 1}{2}} e^{-n + 1} \leq 1 \cdot 3 \cdots (2n - 1) \leq (2n + 1)^{\frac{2n + 1}{2}} \frac{e^{-n + 1}}{3^{3/2}} < \left( \frac{2n + 1}{e} \right)^{\frac{2n + 1}{2}},
\]  
using the fact that \(1 < e < 3\).