A subtle case where $3$ as the parameter gives a simpler result than $2$:

Consider the infinite product

$$\prod\limits_{n=2}^\infty\dfrac{n^k-1}{n^k+1},$$

which converges for all $k>1$. For $k=3$ we can use the factorizations of $a^2\pm b^3$ to telescope the product in terms of rational functions and thus guarantee a rational value, namely $2/3$. For $k=2$ a telescoping is still possible but more complicated, requiring the use of the gamma function and identities involving it to attain the result $\pi/{\sinh(\pi)}$.

A subtle case where $3$ as the parameter gives a simpler result than $2$:

Consider the infinite product

$$\prod\limits_{n=2}^\infty\dfrac{n^k-1}{n^k+1},$$

which converges for all $k>1$. For $k=3$ we can use the factorizations of $a^3\pm b^3$ to telescope the product in terms of rational functions and thus guarantee a rational value, namely $2/3$. For $k=2$ a telescoping is still possible but more complicated, requiring the use of the gamma function and identities involving it to attain the result $\pi/{\sinh(\pi)}$.

A subtle case where $3$ as the parameter gives a simpler result than $2$:

Consider the infinite product

$\prod\limits_{n=2}^\infty\dfrac{n^k-1}{n^k+1},$

which converges for all $k>1$. For $k=3$ we can use the facrorizations of $a^2\pm b^3$ to telescope the product in terms of rational functions and thus guarantee a rational value, namely $2/3$. For $k=2$ a telescoping is still possible but more complicated, requiring the use of the gamma function and identities involvibg it to attain the result $\pi/\sinh(\pi)$.

A subtle case where $3$ as the parameter gives a simpler result than $2$:

Consider the infinite product

$$\prod\limits_{n=2}^\infty\dfrac{n^k-1}{n^k+1},$$

which converges for all $k>1$. For $k=3$ we can use the factorizations of $a^2\pm b^3$ to telescope the product in terms of rational functions and thus guarantee a rational value, namely $2/3$. For $k=2$ a telescoping is still possible but more complicated, requiring the use of the gamma function and identities involving it to attain the result $\pi/{\sinh(\pi)}$.