Show that |1–|3–|5–....|2n−1–––––––>(|n––)n.
|r––|2n−r–––––––|n––>|n––
|1––|2n−1––––––––|n––>|n––
|2––|2n−2––––––––|n––>|n––
|3––|2n−3––––––––|n––>|n––
.
.
.
|2n−1––––––––|1––|n––>|n––
Multiplying the above equation we get .
(|1––|2––|3––...|2n−1––––––––)2|n––n>|n––n
(|1––|2––|3––...|2n−1––––––––)2>|n––2n
(|1––|2––|3––...|2n−1––––––––)>|n––n