# Pulchritude

Calculus Level pending

$\sum_{1\leq n_1<n_2<n_3}\frac{1}{n_1^2 n_2^2 n_3^2}$ If the above expression equals $$\displaystyle \frac{\pi^r}{s}$$, where $$r$$ and $$s$$ are positive integers, what is $$r+s$$?

Here is the summation expanded to the "first" 10 terms:

$$\frac{1}{ 1 ^2\cdot 2 ^2\cdot 3 ^2}+\frac{1}{ 1 ^2\cdot 2 ^2\cdot 4 ^2}+\frac{1}{ 1 ^2\cdot 3 ^2\cdot 4 ^2}+\frac{1}{ 2 ^2\cdot 3 ^2\cdot 4 ^2}+ \frac{1}{ 1 ^2\cdot 2 ^2\cdot 5 ^2}+$$

$$\frac{1}{ 1 ^2\cdot 3 ^2\cdot 5 ^2}+\frac{1}{ 1 ^2\cdot 4 ^2\cdot 5 ^2}+\frac{1}{ 2 ^2\cdot 3 ^2\cdot 5 ^2}+\frac{1}{ 2 ^2\cdot 4 ^2\cdot 5 ^2}+\frac{1}{ 3 ^2\cdot 4 ^2\cdot 5 ^2}+\ldots$$

The sum is taken over all unordered triples $$\{n_1,n_2,n_3\}$$ of distinct positive integers.

Alternatively, this can be represented as $$\displaystyle\sum_{n_3=3}^{\infty}\sum_{n_2=2}^{n_3-1}\sum_{n_1=1}^{n_2-1}\frac{1}{n_1^2 n_2^2 n_3^2}$$.

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