Is $X^2$ Equal to or Greater than 0.04?

[anemone]

Let $X=\dfrac{4}{5}\times\dfrac{6}{7}\times\dfrac{8}{9}\times\cdots\times\dfrac{9998}{9999}$.

State with reason which of the following is true?

I.	$X^2=0.0004$
II.	$X^2\le 0.0004$
III.	$X^2\ge 0.0004$
IV.	$X^2= 0.04$
V.	$X^2\ge 0.04$

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[anemone]

To all members and guests who have read this past week's POTW,

I sincerely apologize for not double checking the inequality signs that I used because the choices of II, III and V should not be weak inequalities but should be strict instead, and I'm truly sorry for making this blunder in last week's POTW.

I've learned a lesson here, I should always double, if not triple check the problem before posting, and I encourage anyone who suspects there is a problem with the wording or presentation of my subsequent POTWs to PM me for clarification and I hope to see more participation in the coming POTWs.

Thanks for reading.

State with reason which of the following is true?

I.	$X^2=0.0004$
II.	$X^2\le 0.0004$ This should be $X^2< 0.0004$
III.	$X^2\ge 0.0004$ This should be $X^2> 0.0004$
IV.	$X^2= 0.04$
V.	$X^2\ge 0.04$ This should be $X^2> 0.04$

No one answered last week's problem (potentially due to the wrong inequality signs that may have led to confusion).

You can find the solution below:

Spoiler

Given $X=\dfrac{4}{5}\times\dfrac{6}{7}\times\dfrac{8}{9}\times\cdots\times\dfrac{9998}{9999}$ and if we let $Y=\dfrac{5}{6}\times\dfrac{7}{8}\times\dfrac{9}{10}\times\cdots\times\dfrac{9999}{10000}$, note that

1. $XY=\dfrac{4}{10000}$

2. $n^2-1<n^2$ this gives $(n+1)(n-1)<n(n)\,\,\,\rightarrow\dfrac{n-1}{n}<\dfrac{n}{n+1}$ which means $X<Y$.

Combining these two observations we can conclude that:

$X^2<\dfrac{4}{10000}=0.0004$ so the answer is B.