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

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

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

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

Hint: $$1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}.$$
• True, but now if we start putting in values of $'n'$ writing $a_n$, there is no single cancellation pattern that can tell us where the limit will go – Abhay Jan 17 at 21:20
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 $$\frac{1}{9}$$.