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.

  • $\begingroup$ Note: for $0 < x < 1$, we have $\log (1-x) < -x$. $\endgroup$ – Daniel Fischer Nov 13 '15 at 16:54
  • $\begingroup$ i have got $$\frac{\Gamma \left(\frac{n+1}{2}\right)}{\sqrt{\pi } \Gamma \left(\frac{n+2}{2}\right)}$$ $\endgroup$ – Dr. Sonnhard Graubner Nov 13 '15 at 16:55
  • $\begingroup$ I'm sorry Daniel Fischer, but I don't know how this could help me. $\endgroup$ – Harald Meisner Nov 13 '15 at 16:58

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 $0$.

  • $\begingroup$ The sum converges? That would mean the limit of the product is not 0. Am I right with this? $\endgroup$ – Harald Meisner Nov 13 '15 at 17:05
  • $\begingroup$ Okay, thank you. I was able to prove it with the hint of Daniel Fischer in the comments to my question. The sum diverges to -$\infty$, hence the product converges to 0. $\endgroup$ – Harald Meisner Nov 13 '15 at 17:31

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).

  • $\begingroup$ Very nice proof $\endgroup$ – Harald Meisner Nov 13 '15 at 17:53

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