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enter image description here I am stuck at this question , any help is appreciated EDIT:the second choice is A < B

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    You can immediately exclude $A=B$, $A=2B$ and $A=3B$ because $A$ has an odd denominator, while $B$ has an even denominator. – 2017-02-25

2 Answers 2

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Render the profuct on the left as

$(2/3)^2(4/5)^2(6/7)^2(8/9)^2...(2014/2015)^2$

Define acomparison product:

$(2/4)(4/6)(6/8)(8/10)...(2014/2016)$

which telescopes to $1/1008$.

Then $(2/3)^2<(2/4)$ because $2\times 4 <3\times 3$. Perform a similar comparison between $(4/5)^2$ and $(4/6)$, $(6/7)^2$ and $(6/8)$, etc.

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    Great solution , thank you sir – 2017-02-25
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Let's generalize and compare $ A_n:=\left(\frac 23\cdot\frac 45\cdot\ldots\cdot \frac{2n}{2n+1}\right)^2$ with $B_n:=\frac1{{n+1}}$. For $n=1$, we have $\left(\frac23\right)^2=\frac49\color{red}<\frac12$, for $n=2$, $\left(\frac23\frac45\right)^2=\frac{64}{225}\color{red}<\frac13$. We suspect that this holds for all $n$. Indeed, using $$\tag1A_{n}=A_{n-1}\cdot\frac{4n^2}{(2n+1)^2} $$ and $$\tag2B_n=B_{n-1}\cdot\frac{n}{n+1} $$ we can make an induction proof provided we know that the factor in $(1)$ is smaller than that in $(2)$. Let's check: $$\frac{4n^2}{(2n+1)^2}- \frac{n}{n+1}=\frac{4n^2(n+1)-n(2n+1)^2}{(n+1)(2n+1)^2}=\frac{-n}{(n+1)(2n+1)^2}<0$$

In summary, $A