Finding limit of $\prod_{t=1}^{n}{\left(1-\frac{2}{(n)(n+1)}\right)^2}$

Question

Let

$$x_n=\left(1-\frac{1}{3}\right)^2\cdot\left(1-\frac{1}{6}\right)^2\cdot\left(1-\frac{1}{10}\right)^2\cdot\left(1-\frac{1}{15}\right)^2\cdots\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2 \quad ,n\geq2 $$

then $\lim_ {n\to\infty} x_n$ is

(1) $\frac{1}{3}$

(2) $\frac{1}{9}$

(3) $\frac{1}{81}$

(4) 0

Answer 1

Your method is wrong: the series

$$\sum_{n=1}^\infty \frac{2}{n(n+1)} = 2\not\to +\infty,$$ therefore you can't conclude that $\ln x_n\to 0$.

Answer 2

One can simplify $\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2=(1-\frac{2}{n+1}+\frac{2}{n})^2=(1-\frac{2}{n+1}+\frac{2}{n})(1+\frac{2}{n+1}-\frac{2}{n})$ and then you get a lot of cancellations in the product.

Answer 3

$(1-\frac{1}{\frac{n(n+1)}{2}})^2 =\frac{(n-1)(n+2)}{n(n+1)}$ which when mulitiplying causes to cancellation and will get a simple ratio,then apply limit