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Hence prove that
P = {2\over 1}{\cdot}{4\over 3}{\cdot}{6\over 5}{\cdots} {k\over k-1}> \sqrt{k+1}.
This shows that the product P=\prod_{i=1}^n{2i\over{2i-1}} tends to infinity as n tends to infinity. Now, using a similar method, show that
Q = {2\over 1}{\cdot}{4\over 3}{\cdot}{6\over 5}{\cdots}{100\over 99}> 12.
Diverging
Age 16 to 18
Challenge Level 
Show that for natural numbers x and y if {x\over y}> 1 then
{x\over y}> {(x+1)\over(y+1)}> 1.
Hence prove that
P = {2\over 1}{\cdot}{4\over 3}{\cdot}{6\over 5}{\cdots} {k\over k-1}> \sqrt{k+1}.
This shows that the product P=\prod_{i=1}^n{2i\over{2i-1}} tends to infinity as n tends to infinity. Now, using a similar method, show that
Q = {2\over 1}{\cdot}{4\over 3}{\cdot}{6\over 5}{\cdots}{100\over 99}> 12.
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