Evaluate X/Y

[anemone]

Let \(\displaystyle X=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2011\cdot2012}\) and \(\displaystyle Y=\frac{1}{1007\cdot2012}+\frac{1}{1008\cdot2011}+\cdots+\frac{1}{2012\cdot1007}\).

Evaluate \(\displaystyle \frac{X}{Y}\).

 

[anemone]

My solution:

Spoiler

\(\displaystyle X=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2011\cdot2012}\)

\(\displaystyle \;\;\;=\sum_{n=1}^{1006} \left( \frac{1}{2n-1}-\frac{1}{2n} \right)\)

\(\displaystyle \;\;\;=\left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+\cdots+\left( \frac{1}{2011}-\frac{1}{2012} \right)\)

Okay, up to this point, we see $X$ and $Y$ aren't closely related so we need to begin to work on $Y$ to gain perspective to see how we should proceed to solve the problem.

\(\displaystyle Y=\frac{1}{1007\cdot2012}+\frac{1}{1008\cdot2011}+\cdots+\frac{1}{2012\cdot1007}\)

\(\displaystyle \;\;\;=\sum_{n=1}^{1006} \frac{1}{3019}\left( \frac{1}{n+1006}+\frac{1}{2013-n} \right)\)

\(\displaystyle \;\;\;=\frac{1}{3019} \sum_{n=1}^{1006} \left( \frac{1}{n+1006}+\frac{1}{2013-n} \right)\)

\(\displaystyle \;\;\;=\frac{1}{3019} \left( \left( \frac{1}{1007}+\frac{1}{2012} \right)+ \left( \frac{1}{1008}+\frac{1}{2012} \right)+\cdots+\left( \frac{1}{2012}+\frac{1}{1007} \right)\right)\)

\(\displaystyle \;\;\;=\frac{2}{3019} \left( \frac{1}{1007}+\frac{1}{1008}+\cdots+\frac{1}{2011}+\frac{1}{2012} \right)\)

Hey, now everything has become so obvious that

\(\displaystyle \left( \frac{1}{1007}+\frac{1}{1008}+\cdots+\frac{1}{2011}+\frac{1}{2012} \right)=\left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+\cdots+\left( \frac{1}{2011}-\frac{1}{2012} \right)\)

and therefore

\(\displaystyle Y=\frac{2X}{3019}\)

\(\displaystyle \frac{X}{Y}=\frac{3019}{2}\)