Prove that 12n lessthanorequalto 1 middot 3 middot 5 2 n 12

Prove that 1/(2n) lessthanorequalto [1 middot 3 middot 5 (2 n - 1)]/(2 middot 4.. ... 2n) whenever n is positive integer.

Solution

PART 1: Show that the statement is true for n = 1

1/2 1/(1.2)
1/2 1/2
1/2 1/2
TRUE

PART 2:
Assume that (135(2n-1))/(246(2n))1/2n is true for n = k.
That is, (135(2k-1))/(246(2k))1/2k

Use this to prove that (135(2n-1))/(246(2n))1/2n is true for n = k+1.
That is, (135(2(k+1)-1))/(246(2(k+1)))...
or (135(2k+1)) / (246(2k+2)) 1/2(k+1)

(135(2k-1)) * (2k+1) / [ (246(2k)) * (2k+2) ] 1/2k * (2k+1) / (2k+2)
(135(2k+1) / (246(2k+2) (2k+1) / (2k+2) * 1/2k)

We can use transitivity (if a < b and b < c, then a < c) by showing that:
if
(135(2k+1) / (246(2k+2) (2k+1) / (2k+2) * 1/2k 1/(2k+2)
and
(2k+1) / (2k+2) * 1/(k+1) 1/(2k+2)
then
(135(2k+1) / (246(2k+2) 1/(2k+2), which we need to prove.

We want to show that
(2k+1) / (2k+2) * 1/2k?? 1/2k
(2k+1)(2k+2) ?? (2k+2)2k
(2k+1) (k+2) ?? (2k+2) (2k)
(2k^2 + 3k + 2) ?? (4k^2 + 4k )
2<2k^2+k
--- Since k is a positive integer, then the last statement is always true. Hence, we can now remove the ??.

Finally, by transitivity, (135(2k+1) / (246(2k+2) 1/2k
Part 2 is now complete.

HENCE, (135(2n-1))/(246(2n))1/2k where n=1,2…

PROVED