Limit Problem 2

Note first that $i^{3} - 1 = (i - 1)(i^{2} + i + 1)$ and $i^{3} + 1 = (i + 1)(i^{2} - i + 1)$ .

Now look at a 3-term expansion of the product starting at any $i \ge 2$ :

$\dfrac{(i - 1)(i^{2} + i + 1)}{(i + 1)(i^{2} - i + 1)}*\dfrac{i((i + 1)^{2} + (i + 1) + 1)}{(i + 2)((i + 1)^{2} - (i + 1) + 1)}*\dfrac{(i + 1)((i + 2)^{2} + (i + 2) + 1)}{(i + 3)((i + 2)^{2} - (i + 2) + 1)}.$

Now note that the first factor in the denominator of the first term cancels the first factor in the numerator of the third term. As we add more terms to the product, similar cancelations will occur with 'period' 2. Next, note that the second factor in the numerator of the first term and the second factor in the denominator of the second term will cancel each other, since

$(i + 1)^{2} - (i + 1) + 1 = i^{2} + 2i + 1 - i = i^{2} + i + 1.$

The same can be said for the second factor in the numerator of the second term and the second factor in the denominator of the third term. As we add more terms to the product, similar cancelations will occur for successive terms.

In this way, as we let $n \rightarrow \infty$ , the product will 'telescope', i.e., all the terms will cancel pairwise except for the first factors of the first two terms of the infinite product, and the second term in the denominator of the first term of the infinite product. Thus the infinite product is equal to

$\dfrac{(2 - 1)*2}{2^{2} - 2 + 1} = \dfrac{2}{3} = \boxed{0.667}$ to $3$ significant figures.

Might be easier to write like this: $\frac{i^{3} -1}{i^{3}+1} = \frac{(i-1)}{(i+1)}\frac{(i+\frac{1}{2})^{2} +\frac{3}{4}}{(i-\frac{1}{2})^{2}+\frac{3}{4}}$ So that we only need to split up the product into two parts - $P= \prod^{n}_{i=2} \frac{i^{3}-1}{i^{3}+1} = \prod_{i=2}^{n} \frac{(i+1)}{(i-1)} \prod_{i=2}^{n} \frac{(i+\frac{1}{2})^{2} +\frac{3}{4}}{(i-\frac{1}{2})^{2}+\frac{3}{4}}$ So that it is (perhaps) quicker to find the telescoping pattern - the first one cancels the lower part of term $i$ with the upper of term $i+2$ , and the second cancels the lower of term $i$ with the upper of term $i+1$ .

$\prod_{i=2}^{n} \frac{(i+1)}{(i-1)} = \frac{2}{n(n+1)}$

$\prod_{i=2}^{n} \frac{(i+\frac{1}{2})^{2} +\frac{3}{4}}{(i-\frac{1}{2})^{2}+\frac{3}{4}} =\frac{(n+\frac{1}{2})^{2}+\frac{3}{4}}{(\frac{3}{2})^{2} +\frac{3}{4} }$

$\lim_{n \rightarrow \infty} P = \lim_{n \rightarrow \infty} \bigg( \frac{2}{3} \times \frac{(n+\frac{1}{2})^{2}+\frac{3}{4}}{n(n+1)} \bigg) = \boxed{\frac{2}{3}}$