<s>[INST] The value of infinite product 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 . 
$$P=\frac{7}{9}\cdot\frac{26}{28}\cdot\frac{63}{65}\cdots\frac{n^3-1}{n^3+1}\cdots.$$
 [/INST] 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 $2/3$.
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