Fraction Sum Reciprocal Product Power

Level 2

If the value of

( 1 1 2 + 1 2 2 + 1 3 2 + … ) ( 1 1 4 + 1 2 4 + 1 3 4 + … ) 1 1 6 + 1 2 6 + 1 3 6 + … \LARGE \displaystyle \frac {\left ( \frac {1}{1^2} + \frac {1}{2^2} + \frac {1}{3^2} + \ldots \right) \left ( \frac {1}{1^4} + \frac {1}{2^4} + \frac {1}{3^4} + \ldots \right) }{ \frac {1}{1^6} + \frac {1}{2^6} + \frac {1}{3^6} + \ldots }

is in the form of a b \frac {a}{b} for coprime positive integers a , b a,b . What is the value of a + b a+b ?


The answer is 11.

Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

Why so large? D:

Let S ( n ) S(n) be the sum of the n-powers of the reciprocals of all positive integers.

It is known that S ( 2 ) = π 2 6 , S ( 4 ) = π 4 90 S(2) = \dfrac{\pi^2}{6}, \; S(4) = \dfrac{\pi^4}{90} and S ( 6 ) = π 6 945 S(6) = \dfrac{\pi^6}{945} .

Thus, the expression can be written as 945 90 × 6 ⇒ 7 4 \dfrac{945}{90 \times 6} \Rightarrow \dfrac{7}{4} . Thus, our desired sum is a + b = 11. a+b = \; \boxed{11.}

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