Proof By Induction

[dagr8est]

For all n >= 1, 1(3)(5)...(2n-1)/[2(4)(6)...(2n)] <= 1/sqrt(2n+1)

Here is what I have done so far.

Base Case(s):
True for n = 1 because 1/2 <= 1/sqrt(3).

Induction Hypothesis:
Assume 1(3)(5)...(2n-1)/[2(4)(6)...(2n)] <= 1/sqrt(2n+1), for some n >= 1.

Induction Step:
1(3)(5)...(2n+1)/[2(4)(6)...(2n+2)]
= 1(3)(5)...(2n-1)/[2(4)(6)...(2n)] (2n+1)/(2n+2)
<= 1/sqrt(2n+1) (2n+1)/(2n+2) [Inductive hypothesis]
= sqrt(2n+1)/(2n+2)

So I know sqrt(2n+1)/(2n+2) <= 1/sqrt(2n+3) for n=2 which is what I'm trying to show but I'm kind of stuck on how to show it from this point.

Any help would be appreciated.

 

[Trebor]

\(\displaystyle \frac{\sqrt{2n+1}}{2n+2} <= \frac{1}{\sqrt{2n+3}}\)
\(\displaystyle \sqrt{(2n+1)(2n+3)} <= 2n+2\)
\(\displaystyle 4n^2+8n+3 <= 4n^2+8n+4\)
\(\displaystyle 3 <= 4\)

You don't need to worry about squaring the 2nd line to get the third line result as the left hand expression in the 2nd line is always positive (in the real domain), so the equaliity holds.