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

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

  • $\begingroup$ 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 $\endgroup$ – Abhay Jan 17 at 21:20
  • $\begingroup$ Are you sure, @learning_maths ? Try writing out several terms in the format Song suggests and look around. $\endgroup$ – Nick Peterson Jan 17 at 21:23
  • $\begingroup$ @learning_maths: you might like to use Song's hint with say the first four terms and see what cancels (just numbers with themselves, not with their multiples) $\endgroup$ – Henry Jan 17 at 21:24

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 $\frac{1}{9}$.


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