The value of infinite product

Question

At a first look I observed that every term of this product is less than 1 so the limit will also be less than 1 but clueless about how to find limit . $$P=\frac{7}{9}\cdot\frac{26}{28}\cdot\frac{63}{65}\cdots\frac{n^3-1}{n^3+1}\cdots.$$

Answer 1

To restate your question: $$ \prod_{n=2}^\infty \frac{n^3-1}{n^3+1} $$ First we begin by expanding the terms: $$ n^3-1 = (n-1)(n^2+n+1) $$ $$ n^3+1 = (n+1)(n^2-n+1) $$ By treating each term as a separate product,and shifting the index of the product, a telescope-like method can be used to simplify the terms into the desired answer. \begin{align*} \prod_{n\ge2}\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)} \\ = \frac{\left(\prod_{n \ge 0} n+1\right)\left(\prod_{n \ge 2} n^2+n+1\right)}{\left(\prod_{n \ge 2} n+1\right)\left(\prod_{n \ge 2} n^2-n+1\right)} \\ = \frac{2\left(\prod_{n \ge 2} n^2+n+1\right)}{\left(\prod_{n \ge 1} (n+1)^2-(n+1)+1\right)} \\ = \frac{2\left(\prod_{n \ge 2} n^2+n+1\right)}{\left(\prod_{n \ge 1} n^2+(2-1)n+(1-1+1)\right)} \\ = \frac{2\left(\prod_{n \ge 2} n^2+n+1\right)}{\left(\prod_{n \ge 1} n^2+n+1\right)} \\ = \frac{2}{3} \end{align*}

Answer 2

Note that $$\ln\left(\frac{n^3-1}{n^3+1}\right)=\ln\left(\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\right)\\=\ln(n-1)-\ln(n+1)+\ln((n+1)^2-(n+1)+1)-\ln(n^2-n+1).$$ Hence, by taking the logarithm of the partial product we obtain a telescoping sum, $$\ln\left(\prod_{n=2}^N \frac{n^3-1}{n^3+1} \right)=\sum_{n=2}^N \ln\left(\frac{n^3-1}{n^3+1}\right)\\=\ln(1)+\ln(2)-\ln(N)-\ln(N+1)+\ln(N^2+N+1)-\ln(3)\\ =\ln(2/3)+\ln\left(\frac{N^2+N+1}{N(N+1)}\right)\stackrel{N\to\infty}{\to} \ln(2/3).$$ Thus the infinite product yields $2/3$.