Who's up to the challenge? 58

Calculus Level 5

$\large \sum_{n=1}^\infty \dfrac{ \zeta(2n)}{(n+1) 2^{4n}} = \dfrac AB - \dfrac {CK}\pi - \dfrac 1D \ln F + \dfrac G{H\pi^I} \zeta(J)$

The equation above holds true for positive integers $A,B,C,D,F,G,H,I$ and $J$ such that $\gcd(A,B) = \gcd(G,H) = 1$ and $F$ is minimized.

Find $A+B+C+D+F+G+H+I+J$ .

Notations :

$\zeta(\cdot)$ denotes the Riemann zeta function .
$K$ denotes the Catalan's constant , $\displaystyle K= \sum_{n=0}^\infty \dfrac{ (-1)^n}{(2n+1)^2} \approx 0.916$

The answer is 55.

1 solution

Mark Hennings
May 29, 2016

There is a typo in the question - it should be $2^{4n}$ in the denominator, not $2^{4k}$ . If we use the formula $\sum_{n=1}^\infty \zeta(2n) z^{2n} \; = \; \tfrac12(1 - \pi z \cot \pi z)$ then the sum is $16\int_0^{\frac14} x(1 - \pi x \cot \pi x)\,dx \; = \; \tfrac12 - \tfrac{4C}{\pi} - \tfrac12\ln2 + \tfrac{35\zeta(3)}{4\pi^2}$ making the answer $1 + 2 + 4 + 2 + 2 + 35 + 4 + 2 + 3 \,=\, \boxed{55}$ .

The question needs a bit of tidying up. Since $\tfrac12\ln2 \; = \; \tfrac{1}{2k}\ln(2^k)$ for any positive integer $k$ , there is more than one solution for $d$ and $f$ as the question is currently expressed.

(+1) The generating function can be proved by first writing $\zeta (2n)$ as the sum definition and then using the result here

Ishan Singh - 5 years ago

Indeed. Or you can look at my post there to see how to deduce the result from "first principles", using complex analysis.

Mark Hennings - 5 years ago

Thanks. I've tidied up the question. @Hummus a , you need to be more careful with your notations.

Pi Han Goh - 5 years ago

Who's up to the challenge? 58

The answer is 55.

1 solution

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