Evaluate $ \prod_{n=3}^\infty ( 1 - \binom{n}{2}^{-1} )$

Question

$$ \prod_{n=3}^\infty \left ( 1 - \frac{1}{\binom{n}{2}} \right)$$

Expanding it out doesn't give any good patterns, I'm totally stuck.

The answer is 1/3 [according to WolframAlpha: https://www.wolframalpha.com/input/?i=product+from+3+to+infinity+(1+-+1%2F((n)+comb+2)+)

Answer 1

Note that we have

$$\begin{align} \prod_{n=3}^N \left(1-\frac{1}{\binom{n}{2}}\right)&=\prod_{n=3}^N \left(1-\frac{2}{n(n-1)}\right)\\\\ &=\prod_{n=3}^N\frac{n^2-n-2}{n(n-1)}\\\\ &=\prod_{n=3}^N \frac{n+1}{n}\frac{n-2}{n-1}\\\\ &=\prod_{n=3}^N \frac{n+1}{n}\prod_{n=3}^N \frac{n-2}{n-1}\\\\ &=\frac{N+1}{3}\frac{1}{N-1} \end{align}$$

Now let $N\to \infty$.

Answer 2

Hint:

$$1-\dfrac1{\binom n2}=\dfrac{n^2-n-2}{n(n-1)}=\dfrac{(n+1)(n-2)}{n(n-1)}=\dfrac{f(n)}{f(n-2)}$$ where $f(m)=\dfrac{m+1}m$

$$\prod_{n=3}^r\dfrac{f(n)}{f(n-2)}=?$$

Answer 3

for the definite product we get $$\prod_{i=1}^m1-\frac{1}{\binom{i}{2}}=1/3\,{\frac {\Gamma \left( m+2 \right) \Gamma \left( m-1 \right) }{ \Gamma \left( m \right) \Gamma \left( m+1 \right) }} $$

Answer 4

Taking the logarithm: (Not necessarily the fastest way here, but in general a good idea for infinite products: you convert these products into series, and we usually know better how to deal with series. Also, logarithms of fractions often lead to telescopic terms.)

$$\begin{align} \ln \prod_{n=3}^\infty \left(1-\binom{n}{2}^{-1}\right) &= \sum_{n=3}^\infty \ln\left(1-\binom{n}{2}^{-1}\right) = \sum_{n=3}^\infty \ln\left(1-\frac{2}{n(n-1)}\right) \\ &= \sum_{n=3}^\infty \ln\left(\frac{n^2-n-2}{n(n-1)}\right) = \sum_{n=3}^\infty \ln\left(\frac{(n-2)(n+1)}{n(n-1)}\right)\\ &= \sum_{n=3}^\infty \left( \ln(n+1) - \ln n-\ln(n-1)+\ln(n-2) \right) \end{align}$$ Now, let us see what telescopes: for $N\geq 3$, \begin{align} \sum_{n=3}^N &\left( \ln(n+1) - \ln n-\ln (n-1)+\ln(n-2) \right) \\&= \sum_{n=3}^N \ln(n+1)-\sum_{n=3}^N \ln n-\sum_{n=3}^N \ln (n-1) +\sum_{n=3}^N \ln(n-2) \\&= \sum_{n=4}^{N+1} \ln n -\sum_{n=3}^N \ln n -\sum_{n=2}^{N-1} \ln n +\sum_{n=1}^{N-2} \ln n \\&= \ln(N+1) - \ln 3 - \ln (N-1) + \ln 1 \\ &= \ln\frac{N+1}{N-1} - \ln 3 \end{align} The first term converges to $\ln 1=0$, so the limit overall is $-\ln 3$. So your original quantity converges, by continuity of $\exp$, to $\boxed{e^{-\ln 3} = \frac{1}{3}}$.

Edit: as mentioned in a comment below: here, clearly one can show the show that the product "telescopes" directly, as Mark Viola does in his answer. However, the point here is to illustrate the use of logarithms to convert products to sums -- it's more general, and is a very good technique to prove convergence/divergence or even obtain asymptotics (when computing the exact value is not easy, or impossible).