MHB Can This Infinite Product Surpass 50?

[anemone]

Here is this week's POTW:

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Is $\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{6}\right)\cdots\left(1+\dfrac{1}{2018}\right)$ greater than, less than or equal to 50?

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[anemone]

Congratulations to Opalg for solving last POTW, which you can find his solution below!

Spoiler

$$\begin{aligned}\left(1+\frac12\right) \left(1+\frac14\right) \left(1+\frac16\right)\cdots \left(1+\frac1{2n}\right) &= \frac32\frac54\frac76 \cdots\frac{2n+1}{2n} \\ &= \frac{(2n+1)!}{4^n(n!)^2} = \frac{(n+1)(2n+1)}{4^n}C_n,\end{aligned}$$ where $C_n = \frac1{n+1}{2n\choose n}$ is the $n$th Catalan number. The Stirling approximation formula for $C_n$ is $C_n \sim \dfrac{4^n}{\sqrt{n^3\pi}}$, and it is accurate to within a factor of about $1/n$. Therefore $$\left(1+\frac12\right) \left(1+\frac14\right) \left(1+\frac16\right)\cdots \left(1+\frac1{2n}\right) \sim \frac{(n+1)(2n+1)}{\sqrt{n^3\pi}}.$$ Put $n = 1009$ to see that $$\left(1+\frac12\right) \left(1+\frac14\right) \left(1+\frac16\right)\cdots \left(1+\frac1{2018}\right) \sim 35.896,$$ correct to about 1 part in 1000. At any rate, it is a lot less than 50.

I will post another solution from other here too for an alternative method to solve the problem.

Spoiler

Let $S=\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{4}\right)\left(1+\dfrac{1}{6}\right)\cdots\left(1+\dfrac{1}{2018}\right)$.

Clearly the following is true:
$S<\left(1+\dfrac{1}{1}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{5}\right)\cdots\left(1+\dfrac{1}{2017}\right)$

Multiply both sides by $S$, we get

$\begin{align*}S^2&<\left(\dfrac{3}{2}\right)\left(\dfrac{5}{4}\right)\left(\dfrac{7}{6}\right)\cdots\left(\dfrac{2019}{2018}\right)\left(\dfrac{2}{1}\right)\left(\dfrac{4}{3}\right)\left(\dfrac{6}{5}\right)\cdots\left(\dfrac{2018}{2017}\right)\\&<2019\\S&<\sqrt{2019}\end{align*}$

Since $2019<2500=50^2$, we get

$S<\sqrt{2019}<50$