Problem 4

$\begin{aligned} a_n & = \displaystyle \prod_{i=2}^n {\frac{i^3-1}{i^3+1}} = \prod_{i=2}^n {\frac{(i-1)(i^2+i+1)}{(i+1)(i^2-i+1)}} \\ & = \dfrac{1\dot{}7}{3\dot{}3} \dot{} \dfrac{2\dot{}13}{4\dot{}7} \dot{} \dfrac{3\dot{}21}{5\dot{}13} \dot{} \dfrac{4\dot{}31}{6\dot{}21} \dot{} \dfrac{5\dot{}43}{7\dot{}31} \dot{} ... \dot{} \dfrac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} \\ & = \left( \dfrac{2\dot{}13}{3\dot{}3\dot{}4} \right) \dot{} \dfrac{3\dot{}21}{5\dot{}13} \dot{} \dfrac{4\dot{}31}{6\dot{}21} \dot{} \dfrac{5\dot{}43}{7\dot{}31} \dot{} ... \dot{} \dfrac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} \\ & = \left( \dfrac{2\dot{}21}{3\dot{}4\dot{}5} \right) \dot{} \dfrac{4\dot{}31}{6\dot{}21} \dot{} \dfrac{5\dot{}43}{7\dot{}31} \dot{} ... \dot{} \dfrac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} \\ & = \left( \dfrac{2\dot{}31}{3\dot{}5\dot{}6} \right) \dot{} \dfrac{5\dot{}43}{7\dot{}31} \dot{} ... \dot{} \dfrac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} \\ & = \dfrac{2(n^2+n+1)}{3n(n+1)} \end{aligned}$

$\Rightarrow \displaystyle \lim_{n \to \infty} {a_n} = \lim_{n \to \infty} {\dfrac{2(n^2+n+1)}{3n(n+1)}} = \lim_{n \to \infty} {\dfrac{2}{3} \left( 1 + \dfrac{1}{n(n+1)} \right)} = \boxed {\dfrac{2}{3}}$

Using the fact that and

We can rewrite the above as :

Now we have two infiite products

Everything except for the 2 cancels out nicely leaving

For the second :

Everything cancels out very nicely again, leaving