Diverging

Problem | Hint | Solution | Printable page |
Stage: 5 Challenge Level: Challenge Level:1

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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Published October 2002.