In exercise 4.16, you proved by induction on n that

Theorem 1 For each n ≥ 1,

∏_{k=1}^{n} \frac{2k − 1}{2k} ≤ \frac{1}{\sqrt{3n + 1}}.  (1)

It follows immediately that

lim_{n→∞} \sqrt{n} ( ∏_{k=1}^{n} \frac{2k − 1}{2k}) ≤ \frac{1}{\sqrt{3}},  (2)

if the limit exists (by no means obvious). In fact, the limit does exist and we have the following result :

Theorem 2

lim_{n→∞} \sqrt{n} ( ∏_{k=1}^{n} \frac{2k − 1}{2k}) = \frac{1}{\sqrt{π}} .  (3)

Proof : Note that

∏_{k=1}^{n} 2k = 2^{n}n!

Multiplying the product in (2) above and below by this number one sees that

∏_{k=1}^{n} \frac{2k − 1}{2k} = \frac{(2n)!}{2^{2n}(n!)^{2}} .  (4)

The proof is now completed by applying a famous formula due to Stirling. I know of no easy proof of this result, so I will just state it. In fact, all known proofs seem to use techniques from a branch of mathematics called complex analysis. which is the theory of differentiable functions of one complex variable. Maybe you’ll take the course some time !

Stirling’s formula

lim_{n→∞} \frac{n!}{\sqrt{2πn} n^{n}e^{−n}} = 1.  (5)

Now back to the proof of Theorem 2. From (4) and (5) we obtain

LHS of ( ) = lim_{n→∞} \sqrt{n} · \frac{\sqrt{2π(2n)} (2n)^{2n}e^{−2n}}{2^{2n} (\sqrt{2πn} n^{n}e^{−n})^{2}} = \frac{1}{\sqrt{π}} ,

as required. 2