Change in Harmony

Calculus Level 5

∑ n = 1 ∞ H n ( 2 ) ( n + 1 ) 2 = π A B \sum _{ n=1 }^{ \infty }{ \dfrac { { H }_{ n }^{ \left( 2 \right) } }{ { \left( n+1 \right) }^{ 2 } } } =\dfrac { { \pi }^{ A } }{ B }

If the equation above holds true for positive integers A A and B B , find A + B . A+B.

Notation : H n ( m ) H_n^{(m)} denotes the generalized harmonic number, H n ( m ) = ∑ k = 1 n 1 k m \displaystyle H_n^{(m)} = \sum_{k=1}^n \dfrac1{k^m} .

Inspiration .


The answer is 124.

Aditya Kumar by Aditya Kumar , 22, India

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1 solution

Mark Hennings
Feb 26, 2016

The sum is ∑ n = 1 ∞ H n ( 2 ) ( n + 1 ) 2 = ∑ n > m ≥ 1 1 m 2 n 2 = 1 2 ∑ m ≠ n 1 m 2 n 2 = 1 2 [ ( ∑ n = 1 ∞ 1 n 2 ) 2 − ∑ n = 1 ∞ 1 n 4 ] = 1 2 [ ζ ( 2 ) 2 − ζ ( 4 ) ] = 1 120 π 4 \begin{array}{rcl} \displaystyle \sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2} \; = \; \sum_{n > m \ge 1} \frac{1}{m^2n^2} & = & \displaystyle\tfrac12\sum_{m \neq n} \frac{1}{m^2n^2} \; = \; \frac12\left[\left(\sum_{n=1}^\infty \frac{1}{n^2}\right)^2 - \sum_{n=1}^\infty \frac{1}{n^4}\right] \\ & = & \tfrac12\big[\zeta(2)^2 - \zeta(4)\big] \; =\; \tfrac{1}{120}\pi^4 \end{array} making the answer 4 + 120 = 124 4 + 120 = \boxed{124} .

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