Can you prove the inequality for positive integers?

[Ackbach]

My apologies for being late this week. Here is this week's POTW:

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

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[Ackbach]

Re: Problem Of The Week # 225 - Jul 21, 2016

This was Problem B-2 in the 1996 William Lowell Putnam Mathematical Competition.

Congratulations to Opalg for his correct (of course) solution. He appeared to be on a roll this week, answering every single level of POTW! Here is his solution:

Spoiler

Let $f(x) = x(\ln x - 1)$. Notice that $f(x)$ is the indefinite integral of $\ln x$. Notice also that $f(1) = -1 < 0$; and $f(3) = 3(\ln3-1) > 0$ (because $3>e$ and so $\ln3 > 1$).

Therefore \(\displaystyle \int_1^{2n-1}\ln x\,dx = f(2n-1) - f(1) > f(2n-1) = (2n-1)\bigl(\ln(2n-1) - 1\bigr).\) Now form the upper Riemann sum for this integral, using subintervals of length $2$ on the interval $[1,2n-1].$ Since $\ln x$ is an increasing function, its supremum on each subinterval will be its value at the right-hand endpoint of the subinterval. The upper Riemann sum is therefore $2\bigl(\ln3 + \ln5 +\ldots + \ln(2n-1)\bigr).$ The upper Riemann sum is larger than the integral, so that $$(2n-1)\bigl(\ln(2n-1) - 1\bigr) < \int_1^{2n-1}\ln x\,dx < 2\bigl(\ln3 + \ln5 +\ldots + \ln(2n-1)\bigr).$$ Adding $\ln 1\;(=0)$ makes no difference to the sum, so $\dfrac{2n-1}2\bigl(\ln(2n-1) - 1\bigr) < \ln1 + \ln3 + \ln5 +\ldots + \ln(2n-1).$ Now take the exponential of both sides to get $$\Bigl(\dfrac{2n-1}e\Bigr)^{(2n-1)/2} < 1\cdot 3 \cdot 5 \cdots (2n-1).$$

The proof for the other inequality is essentially the same. This time, use the interval $[3,2n+1].$ Again, divide it into subintervals of length 2, and form the lower Riemann sum for \(\displaystyle \int_3^{2n+1}\ln x\,dx = f(2n+1) - f(3) < f(2n+1) = (2n+1)\bigl(\ln(2n+1) - 1\bigr)\) (because $f(3) > 0$). The lower Riemann sum this time is $2\bigl(\ln1 + \ln3 + \ln5 +\ldots + \ln(2n-1)\bigr).$ It is less than the integral, and so $$2\bigl(\ln1 + \ln3 + \ln5 +\ldots + \ln(2n-1)\bigr) < \int_3^{2n+1}\ln x\,dx < (2n+1)\bigl(\ln(2n+1) - 1\bigr).$$ Therefore $\ln1 +\ln3 + \ln5 +\ldots + \ln(2n-1) < \dfrac{2n+1}2\bigl(\ln(2n+1) - 1\bigr).$ As previously, exponentiate, to get $$1\cdot 3 \cdot 5 \cdots (2n-1) < \Bigl(\dfrac{2n+1}e\Bigr)^{(2n+1)/2}.$$