Finding limit of a sequence in product form

Question

\begin{equation} \prod_{n=2}^{\infty} \left (1-\frac{2}{n(n+1)} \right )^2 \end{equation}

I need to find limit for the following product..answer is $\frac{1}{9}$.

I have tried cancelling out but can't figure out.

Its a monotonically decreasing sequence so will converge to its infimum.. how to find the infimum?

Answer 1

$$1-\frac2{n(n+1)}=\frac{(n+2)(n-1)}{n(n+1)}=\frac{n-1}n\cdot\frac{n+2}{n+1}$$ waiting for a lot of cancellations

$$\prod_{n=2}^r\left(1-\frac2{n(n+1)}\right)=\left(\prod_{n=2}^r\frac{n-1}n\right)\left(\prod_{n=2}^r\frac{n+2}{n+1}\right)=\frac1r\cdot\frac{r+2}3$$

Now set $r\to\infty$

Answer 2

Hint:

You can write it as $$e^{2\sum_{n=2}^{\infty}f\left(n\right)}$$

Note that $$\log\left(1-\dfrac{2}{n\left(n+1\right)}\right)=\log\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)=\log\left(n-1\right)+\log\left(n+2\right)-\log n-\log\left(n+1\right)$$

Answer 3

The product is equal to $\left(\frac{2-1}{2+1}\frac{n+2}{n}\right)^2$ and the limit is 1/9.