Harmony Squared

Calculus Level 5

A B = ( 1 2 ) 2 + ( 1 3 + 1 6 ) 2 + ( 1 4 + 1 8 + 1 12 ) 2 + ( 1 5 + 1 10 + 1 15 + 1 20 ) 2 + ( 1 1 ) 2 + ( 1 2 + 1 4 ) 2 + ( 1 3 + 1 6 + 1 9 ) 2 + ( 1 4 + 1 8 + 1 12 + 1 16 ) 2 + \dfrac{A}{B}=\dfrac{\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)^2+\left(\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}\right)^2+\left(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{20}\right)^2+\cdots}{\left(\dfrac{1}{1}\right)^2+\left(\dfrac{1}{2}+\dfrac{1}{4}\right)^2+\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{9}\right)^2+\left(\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{16}\right)^2+\cdots}

The equation above holds true for positive coprime integers A A and B B . Find the arithmetic mean of A A and B B .


The answer is 14.

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1 solution

Chew-Seong Cheong
Feb 26, 2018

S A = ( 1 2 ) 2 + ( 1 3 + 1 6 ) 2 + ( 1 4 + 1 8 + 1 12 ) 2 + = k = 1 H k 2 ( k + 1 ) 2 = s h ( 2 , 2 ) where s h ( m , n ) is Euler sum. = 11 360 π 4 See note 1. \begin{aligned} S_A & = \left(\frac 12\right)^2 + \left(\frac 13 + \frac 16 \right)^2 + \left(\frac 14 + \frac 18 + \frac 1{12} \right)^2 + \cdots \\ & = \sum_{k=1}^\infty H_k^2(k+1)^{-2} \\ & = \color{#3D99F6} s_h (2,2) & \small \color{#3D99F6} \text{where }s_h (m,n) \text{ is Euler sum.} \\ & = \color{#3D99F6} \frac {11}{360} \pi^4 & \small \color{#3D99F6} \text{See note 1.} \end{aligned}

S B = ( 1 1 ) 2 + ( 1 2 + 1 4 ) 2 + ( 1 3 + 1 6 + 1 9 ) 2 + = k = 1 H k 2 k 2 See note 2. = s h ( 2 , 2 ) + 2 s h ( 1 , 3 ) + ζ ( 4 ) = 11 360 π 4 + 2 × π 4 360 + π 4 90 See note. = 17 360 π 4 \begin{aligned} S_B & = \left(\frac 11\right)^2 + \left(\frac 12 + \frac 14 \right)^2 + \left(\frac 13 + \frac 16 + \frac 19 \right)^2 + \cdots \\ & = \sum_{k=1}^\infty H_k^2k^{-2} & \small \color{#3D99F6} \text{See note 2.} \\ & = s_h (2,2) + {\color{#3D99F6}2s_h(1,3)} + \zeta (4) \\ & = \frac {11}{360}\pi^4 + {\color{#3D99F6}2\times \frac {\pi^4}{360}} + \frac {\pi^4}{90} & \small \color{#3D99F6} \text{See note.} \\ & = \frac {17}{360} \pi^4 \end{aligned}

Therefore, S A S B = 11 17 \dfrac {S_A}{S_B} = \dfrac {11}{17} , then A + B 2 = 11 + 17 2 = 14 \dfrac {A+B}2 = \dfrac {11+17}2 = \boxed{14} .


See reference: Euler sum

Note 1: s h ( 2 , 2 ) = 3 2 ζ ( 4 ) + 1 2 [ ζ ( 2 ) ] 2 = 11 360 π 4 . . . ( 28 ) \ \displaystyle \small s_h (2,2) = \frac 32 \zeta(4) + \frac 12 [\zeta(2)]^2 = \frac {11}{360}\pi^4 \quad ...(28)

Note 2: k = 1 H k 2 k n = s h ( 2 , n ) + 2 s h ( 1 , n + 1 ) + ζ ( n + 2 ) . . . ( 7 ) \ \displaystyle \small \sum_{k=1}^\infty H_k^2k^{-n} = s_h (2,n) + 2s_h(1,n+1) + \zeta (n+2) \quad ...(7)

Note 3: s h ( 1 , n ) = 1 2 n ζ ( n + 1 ) 1 2 k = 1 n 2 ζ ( n k ) ζ ( k + 1 ) . . . ( 21 ) s h ( 1 , 3 ) = 3 2 ζ ( 4 ) 1 2 ζ ( 2 ) ζ ( 2 ) = π 4 360 \ \displaystyle \small s_h (1,n) = \frac 12 n \zeta(n+1) - \frac 12 \sum_{k=1}^{n-2}\zeta(n-k)\zeta(k+1) \quad ... (21) \\ \ \displaystyle \small \implies s_h(1,3) = \frac 32 \zeta(4) - \frac 12 \zeta(2)\zeta(2) = \frac {\pi^4}{360}

What are the proofs involved?

diwakar kumar - 3 years, 3 months ago

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Ask Euler. Refer to the link: Euler Sum .

Chew-Seong Cheong - 3 years, 3 months ago

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