Pulchritude

Calculus Level 5

∑ 1 ≤ n 1 < n 2 < n 3 1 n 1 2 n 2 2 n 3 2 \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 π r s \displaystyle \frac{\pi^r}{s} , where r r and s s are positive integers, what is r + s r+s ?

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

1 1 2 ⋅ 2 2 ⋅ 3 2 + 1 1 2 ⋅ 2 2 ⋅ 4 2 + 1 1 2 ⋅ 3 2 ⋅ 4 2 + 1 2 2 ⋅ 3 2 ⋅ 4 2 + 1 1 2 ⋅ 2 2 ⋅ 5 2 + \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}+

1 1 2 ⋅ 3 2 ⋅ 5 2 + 1 1 2 ⋅ 4 2 ⋅ 5 2 + 1 2 2 ⋅ 3 2 ⋅ 5 2 + 1 2 2 ⋅ 4 2 ⋅ 5 2 + 1 3 2 ⋅ 4 2 ⋅ 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 } \{n_1,n_2,n_3\} of distinct positive integers.

Alternatively, this can be represented as ∑ n 3 = 3 ∞ ∑ n 2 = 2 n 3 − 1 ∑ n 1 = 1 n 2 − 1 1 n 1 2 n 2 2 n 3 2 \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} .


The answer is 5046.

Kerry Soderdahl by Kerry Soderdahl , 23, USA

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1 solution

Aditya Kumar
Mar 20, 2016

The summation involves multiple zeta values. ∑ n 1 > n 2 > n 3 ≥ 1 1 n 1 2 n 2 2 n 3 2 = ζ ( 2 , 2 , 2 ) \sum _{ { n }_{ 1 }>{ n }_{ 2 }>{ n }_{ 3 }\ge 1 }^{ }{ \frac { 1 }{ { n }_{ 1 }^{ 2 }{ n }_{ 2 }^{ 2 }{ n }_{ 3 }^{ 2 } } } =\zeta \left( 2,2,2 \right)

Now, by Newton's identities, we can write it as: ζ ( 2 , 2 , 2 ) = 3 16 ζ ( 6 ) \zeta \left( 2,2,2 \right) =\frac { 3 }{ 16 } \zeta \left( 6 \right)

Hence, we get: ∑ n 1 > n 2 > n 3 ≥ 1 1 n 1 2 n 2 2 n 3 2 = π 6 5040 \sum _{ { n }_{ 1 }>{ n }_{ 2 }>{ n }_{ 3 }\ge 1 }^{ }{ \frac { 1 }{ { n }_{ 1 }^{ 2 }{ n }_{ 2 }^{ 2 }{ n }_{ 3 }^{ 2 } } }=\frac{\pi^6}{5040}

Keep on posting such problems!

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You might be interested in reading Jake Lai's comment under Mark Henning's solution .

Pi Han Goh - 5 years, 2 months ago

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While solving the problem, I used that. While posting I posted a different one as I wanted others to learn about Zagier’s conjecture.

Aditya Kumar - 5 years, 2 months ago

You can also use Newton's identities, and derive that

ζ ( 2 , 2 , 2 ) = 1 3 ( ζ ( 6 ) − ζ ( 2 ) ζ ( 4 ) + 1 2 ( ζ ( 2 ) 3 − ζ ( 2 ) ζ ( 4 ) ) ) \zeta(2,2,2)=\frac{1}{3}\left(\zeta(6)-\zeta(2)\zeta(4)+\frac{1}{2}\left(\zeta(2)^3-\zeta(2)\zeta(4)\right)\right)

Julian Poon - 5 years, 2 months ago

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Can you show us your working? Thanks.

Pi Han Goh - 5 years, 2 months ago

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Through newton's identities

ζ ( 4 ) = ζ ( 2 ) 2 − 2 ζ ( 2 , 2 ) \zeta(4)=\zeta(2)^2 - 2\zeta(2,2)

ζ ( 2 , 2 ) = 1 2 ( ζ ( 2 ) 2 − ζ ( 4 ) ) \zeta(2,2)=\frac{1}{2}\left(\zeta(2)^2-\zeta(4)\right)

Again through newton identities

ζ ( 2 , 2 , 2 ) = ζ ( 4 ) ζ ( 2 ) − ζ ( 2 , 2 ) ζ ( 2 ) + 3 ζ ( 2 , 2 , 2 ) \zeta(2,2,2)=\zeta(4)\zeta(2) - \zeta(2,2)\zeta(2) + 3\zeta(2,2,2)

Putting it all together and solving for ζ ( 2 , 2 , 2 ) \zeta(2,2,2) gives the above result.

This is similar to what Mark Henning did on Jake Lai's Problem

Julian Poon - 5 years, 2 months ago

In one pdf i had seen this under Zagier’s conjecture. My method was the same.

Aditya Kumar - 5 years, 2 months ago

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