Limit of $a_n=(1-\frac13)^2\cdot(1-\frac16)^2\ldots(1-\frac{1}{\frac{(n)(n+1)}{2}})^2 \; \;\forall n \geq 2$

Question

$$a_n = \left(1-\frac13\right)^2\cdot\left(1-\frac16\right)^2\ldots\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2 \; \;\forall n \geq 2$$

I have no idea how to solve this, however I will give it a try using Cauchy's second theorem... the following is probably wrong since all terms will depend on $'n'$ after taking $nth$ root

$$\lim_{n\rightarrow\infty} a_n = \left(1-\frac13\right)^2\cdot\left(1-\frac16\right)^2\cdot\ldots\cdot\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2 = \lim_{n\rightarrow\infty}\left[\left(1-\frac13\right)^{2n}\cdot\left(1-\frac16\right)^{2n}\cdot\ldots\cdot\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2n}\right]^{\frac1n} = \lim_{n\rightarrow\infty}\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2n}=1 $$

Answer 1

Hint: $$ 1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}. $$

Answer 2

One can generalise $a_n = \frac{(n+3)^2}{9\cdot (n+1)^2}$ by considering the cancellation of each additional term during multiplication. By taking the limit as $n \to \infty$ we see that the resulting value is $\frac{1}{9}$.

Answer 3

Note that: $1- \frac{2}{n(n+1)}= \frac{(n-1)(n+2)}{n(n+1)}$

So we have : \begin{align} & x_n= \left [\frac{1\cdot 4}{2\cdot 3} \cdot \frac{2\cdot 5}{3\cdot 4}\cdot \frac{3\cdot 6}{4\cdot 5} \cdot ...\\ \cdot \frac{(n-2)(n+1)}{(n-1)n} \cdot\frac{(n-1)(n+2)}{n(n+1)} \right]^2 \\ & = \left (\frac{n+2}{3n} \right)^2 \\ & = \frac{1}{9}\left (1+\frac{4}{n}+\frac{4}{n^2} \right) \end{align} So $\displaystyle \lim_{x \to \infty} x_n= \frac{1}{9}$