limit of $\lim_{n \to \infty}\prod\limits_{k=0}^{\frac{n}{2}-1} \left(1-\frac{1}{2k+2} \right)$

Question

I want to calculate the limit of

$$\lim_{n \to \infty}\prod\limits_{k=0}^{\frac{n}{2}-1} \left(1-\frac{1}{2k+2} \right)$$

I think, that it's $0$, but I don't know how to prove this. I can't even say what I've tried so far since I really just looked at a few values with Wolfram Alpha. I've never worked with infinite products before.

Answer 1

Note that your product consist of only positive numbers, so $\prod_{k=0}^{n/2-1} (1 - \frac{1}{2k+2})>0$ for all $n$.
Now, you know that $e^{\ln{a}} =a$ for all $a>0$, so you get $$ \prod_{k=0}^{\frac{n}{2}-1} (1 - \frac{1}{2k+2}) = \exp \left( \ln \prod_{k=0}^{\frac{n}{2}-1} (1 - \frac{1}{2k+2})\right) = \exp \left( \sum_{k=0}^{\frac{n}{2}-1} \ln (1 - \frac{1}{2k+2})\right)$$

Now, if you can show that the sum convergences, you will have a non-zero limit as $\lim_{ n \to \infty} e^{a_n} = e^{ \lim_{n \to \infty} a_n}$.
Otherwise, if the sum diverge to $\infty$, you find that the product also diverges to $\infty$.
If on the other hand the sum diverges to $-\infty$, you find that the product converges to $0$.

Answer 2

Using the double factorial notation (see here) we have $$\prod_{k=0}^{N}\left(1-\frac{1}{2k+2}\right)=\prod_{k=0}^{N}\left(2k+1\right)\prod_{k=0}^{N}\left(2k+2\right)^{-1}=\frac{\left(2N+1\right)!!}{\left(2N+2\right)!!} $$ and so using the identities $$\left(2N+1\right)!!=\frac{\left(2N+1\right)!}{2^{N}N!} $$ and $$\left(2N+2\right)!!=2^{N+1}\left(N+1\right)! $$ we have $$\prod_{k=0}^{N}\left(1-\frac{1}{2k+2}\right)=\frac{\left(2N+1\right)!}{2^{2N+1}N!\left(N+1\right)!}\rightarrow0$$ (for the last limit you can use the Stirling's approximation).