Prove $ \left(\frac{2n-1}{e}\right)^{\frac{2n-1}{2}} <\quad 1\times3\times\dotsb\times(2n-1) \quad<\quad \left(\frac{2n+1}{e}\right)^{\frac{2n+1}{2}}$

Question

While browsing through past years Putnam problems, I found one that stumped me giving no idea of approach.

Prove $$ \left(\frac{2n-1}{e}\right)^{\frac{2n-1}{2}} <\quad 1\times3\times5\times\dotsb\times(2n-1) \quad<\quad \left(\frac{2n+1}{e}\right)^{\frac{2n+1}{2}}$$

Any help will be appreciated.

Answer 1

Following the comment by Wojowu, interpret sums as upper and lower Riemann sums $$ \int_1^{2n-1}\ln(x)\,dx\le\sum_{k=2}^n\ln(2k-1)·2=\sum_{k=1}^n\ln(2k-1)·2\le \int_3^{2n+1}\ln(x)\,dx $$ and as $(x\ln x)'=\ln x+1$, one can easily compute the integral values.