pi's and products

$A=\prod _{ n=2 }^{ \infty }{ \left( 1-\frac { 1 }{ { n }^{ 3 } } \right) } \\ \quad =\prod _{ n=2 }^{ \infty }{ \left( \frac { (n-1)({ n }^{ 2 }+n+1) }{ { n }^{ 3 } } \right) } \\ B=\prod _{ n=2 }^{ \infty }{ \left( 1+\frac { 1 }{ n(n+1) } \right) } \\ \quad =\prod _{ n=2 }^{ \infty }{ \left( \frac { ({ n }^{ 2 }+n+1) }{ n(n+1) } \right) } \\ \frac { A }{ B } =\frac { \cfrac { 1 }{ 2 } }{ \cfrac { 3 }{ 2 } } \quad \quad \quad \quad \quad \quad [the\quad other\quad terms\quad would\quad cancel]\\ \quad \frac { A }{ B } =\frac { 1 }{ 3 } \quad \Longrightarrow m=1;n=3\\ \therefore \quad 100m+n=103\\$

First, we write out the given fraction.

$\frac{A}{B} = \frac{\prod_{a=2}^{\infty} 1-\frac{1}{a^{3}}}{\prod_{b=1}^{\infty} 1+\frac{1}{b(b+1)}}$

Since these are products, we can combine the numerator and denominator to give a single product from $n = 2$ to $\infty$ by factoring out the first term of $B$ .

$\frac{A}{B} = \frac{2}{3} \prod_{n=2}^{\infty} \frac{1-\frac{1}{n^{3}}}{1+\frac{1}{n(n+1)}}$

We do further simplifying.

$\frac{A}{B} = \frac{2}{3} \prod_{n=2}^{\infty} \frac{(n^{3}-1)(n(n+1))}{n^{3}(n(n+1)+1)}$

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

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

From this, we note that terms cancel out until we are left with $\frac{n-1}{n} = \frac{1}{2}$ .

Finally,

$\frac{A}{B} = \frac{2}{3}(\frac{1}{2}) = \frac{1}{3}$ .

$m = 1, n = 3 \rightarrow 100m+n = \boxed{103}$