Diverging
Stage: 5 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
$$P = {2\over 1}{\bf .}{4\over 3}{\bf .}{6\over 5}{\bf ...}
{k\over k-1}> \sqrt{k+1}.$$
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
$$Q = {2\over 1}{\bf .}{4\over 3}{\bf .}{6\over 5}{\bf
...}{100\over 99}> 12.$$
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