Funny if you find the key...

Here is the key:

(1) Multiply the above expression by:

$\left( 1 + \frac11 \right) \left( 1 + \frac13 \right) \left( 1 + \frac15 \right)\cdots\left( 1 + \frac{1}{2017} \right)$

to get the whole thing equal to $2019 .$

(2) Now it is very easy to prove that

$\left( 1 + \frac12 \right) \left( 1 + \frac14 \right) \left( 1 + \frac16 \right)\cdots\left( 1 + \frac{1}{2018} \right) < \left( 1 + \frac11 \right) \left( 1 + \frac13 \right) \left( 1 + \frac15 \right)\cdots\left( 1 + \frac{1}{2017} \right)$

(3) Thus:

$\left( 1 + \frac12 \right) \left( 1 + \frac14 \right) \left( 1 + \frac16 \right)\cdots\left( 1 + \frac{1}{2018} \right) < \sqrt{2019} \\ \color{#3D99F6} \left( 1 + \frac12 \right) \left( 1 + \frac14 \right) \left( 1 + \frac16 \right)\cdots\left( 1 + \frac{1}{2018} \right) < 50$

Let the number be $X$ and then we express $\begin{aligned} X & = \dfrac{3\times 5\times 7\times \cdots \times 2017\times 2019}{2\times 4 \times 6\times \cdots \times 2016\times 2018}\\ & = 2019\underbrace{\left(\dfrac{2017!!}{2018!!}\right)}_{\text{note }} = 2019\underbrace{\left(\dfrac{2018!}{2^{2018} \cdot (1009!)^2}\right)}_{\text{note}}= \dfrac{2019}{2^{2018}}\binom{2018}{1009}\end{aligned}$ We have that asymptotic form for central binomial $\binom{2n}{n}\approx \dfrac{4^n}{\sqrt{\pi n}}$ Setting $n =1009$ which follows as $X = \dfrac{2019}{2^{2018}}\cdot \dfrac{4^{1009}}{\sqrt{1009\pi}}=\dfrac{2019}{\sqrt{1009\pi } }\approx 35.860 < 50$

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Note: $(2n-1)!!= \frac{(2n)!}{2^n\cdot n!}$ and $(2n)!! = 2^n\cdot n!$ .

Let the given product be $\displaystyle P = \prod_{k=1}^{1009} \left(1+\frac 1{2k}\right) = \exp \left(\sum_{k=1}^{1009} \ln \left(1+\frac 1{2k}\right)\right)$ , where $\exp(x) = e^x$ . Let $\displaystyle S = \sum_{k=1}^{1009} \ln \left(1+\frac 1{2k}\right)$ and we can estimate it with the following integral $I$ which is slightly larger than $S$ .

$\begin{aligned} I & = \int_{0.5}^{1009.5} \ln \left(1+\frac 1{2x}\right) dx = \int_{0.5}^{1009.5} \ln \left(\frac {2x+1}{2x} \right) dx \\ & = \int_{0.5}^{1009.5} \bigg(\ln (2x+1) - \ln (2x)\bigg) dx \\ & = \int_{0.5}^{1009.5} \left(\ln 2 + \ln \left(x+ \frac 12 \right) - \ln 2 - \ln x \right) dx \\ & = \left[ \left(x+ \frac 12 \right) \ln \left(x+ \frac 12 \right) - x\ln x \right]_{0.5}^{1009.5} \\ & \approx 3.612155432 \end{aligned}$

Therefore, $P = e^S < e^I \approx e^{3.612155432} \approx 37.046 \boxed {< 50}$ .