Denote the left-hand side of the inequality A:

And introduce its nemesis B:

Factor by factor, the fractions in B exceed those in A:

From this it follows that A < B. Note that, due to the choice of B, in the product AB most of the terms cancel out: AB = 1/100. From here,

which, with one additional step, proves (1).

This proof suggests that (1) is in fact just a special case of a more general inequality

(2)	1/2·3/4·5/6· ... ·(2n-1)/2n < 1/(2n),

whose proof is a slight modification of the above with A and B defined as

As we shall see shortly, (1) and (2) are quite weak: A(n) has a much better bound, viz.

(3)	A(n) < 1/(3n+1).

To remind,

and we wish to prove (3): A(n) < 1/(3n+1). For n = 1, we have

A(1) = 1/2 = 1/(3·1+1).

But already for n = 2,

A(2) = 1/2·3/4 = 3/8 < 1/7 = 1/(3·2+1),

because upon squaring 9/64 < 1/7, for 7·9 = 63 < 64. Thus let's proceed with the inductive step and assume that (3) holds for n = k:

(4)	A(k) < 1/(3k+1).

We are going to prove that, for n = k+1, (3) also holds

(5)	A(k+1) < 1/(3(k+1)+1) = 1/(3k+4).

Since A(k+1) = A(k)·(2k+1)/(2k+2), (4) implies

(6)	A(k+1) < (2k+1)/(2k+2)·1/(3k+1).

Now square the right hand side in (6):

(7)	[(2k+1)/(2k+2)·1/(3k+1)]2 = (2k+1)2 / ((2k+2)2(3k+1))   = (2k+1)2 / (12k3 + 28k2 + 20k + 4)   = (2k+1)2 / [(12k3 + 28k2 + 19k + 4) + k]   = (2k+1)2 / [(2k+1)2(3k+4) + k]   < (2k+1)2 / [(2k+1)2(3k+4)]   = 1 / (3k+4),

which is exactly the right-hand side of (5) and proves (6).