The floor of a product of fractions

Question

Evaluate: $ \displaystyle \Bigg \lfloor \prod_{n=0}^{248} \frac{33+8n}{29+8n} \Bigg \rfloor= \Bigg \lfloor \frac{33}{29} \times \frac{41}{37} \times \frac{49}{45} \times\ ...\ \times \frac{2009}{2005} \times \frac{2017}{2013} \Bigg \rfloor $

I tried to tackle this problem by first trying to place bounds on the product itself using various inequalities such as AM-GM and Cauchy-Schwarz. I did get somewhere and I started to get hopeful but it didn't quite work out. It seemed as if I could place an upper bound but not a lower bound. Maybe Rearrangement/Muirhead/Chebyshev/Holder's inequalities could help but I haven't tried those yet, as a matter of fact I don't think I remember many of them either. Any help on this will be highly appreciated. Thanks.

Answer 1

$$\prod_{n=0}^{248}\frac{33+8n}{29+8n}=\prod_{n=0}^{248}\left(1+\frac4{8n+29}\right)\simeq\prod_{n=0}^{248}\left(1+\frac4{8n+28}\right)=\prod_{n=0}^{248}\left(1+\frac1{2n+7}\right)=$$

$$=\prod_{n=0}^{248}\left[1+\frac1{2(n+3)+1}\right]=\prod_{k=3}^{251}\left(1+\frac1{2k+1}\right)=\prod_{k=3}^{251}\frac{2k+2}{2k+1}=\prod_{k=3}^{251}\frac{k+1}{k+\frac12}=\frac5{16}\prod_{k=0}^{251}\frac{k+1}{k+\frac12}$$

$$=\frac5{16}\cdot\frac{\Gamma(253)}{\Gamma\left(252+\frac12\right)}\cdot\sqrt\pi=\frac5{16}\cdot\frac{252!}{504!\cdot2^{-504}\cdot(252!)^{-1}\cdot\sqrt\pi}\cdot\sqrt\pi=\frac5{16}\cdot2^{504}\cdot{504\choose252}^{-1}\simeq$$

$$\simeq\frac5{16}\cdot2^{504}\cdot\frac{\sqrt{252\pi}}{2^{504}}=\frac5{16}\cdot6\sqrt{7\pi}=\frac{15}8\sqrt{7\pi}\simeq8.8$$

In the above calculations, we've used the famous multiplication formula for the Gamma function, as well as the well-known approximation for the central binomial coefficient.

Answer 2

I assume you cannot use a calculator, but can do simple division by hand and spot whether a number is between $8^2$ and $9^2$.

If you think $\dfrac{x+2}{x-2} \le \sqrt{\dfrac{x+4}{x-4}}$ is a reasonable approximation then your calculation becomes less than but close to $\displaystyle \Bigg \lfloor \sqrt{ \frac{2019}{27} } \Bigg \rfloor$. The approximation is quite good here: it gives $\lfloor 8.6474\ldots \rfloor$ rather than the correct $\lfloor 8.6357\ldots \rfloor$.

You can do something similar for a lower bound: for example $\dfrac{x+4}{x} \ge \sqrt{\dfrac{x+8}{x}}$, so a lower bound is $\displaystyle \Bigg \lfloor \sqrt{ \frac{2021}{29} } \Bigg \rfloor = \lfloor 8.3480\ldots \rfloor$.

So your answer is $8$.