A Squared Riemann Zeta

Calculus Level 5

$\large \sum_{n=1}^\infty \dfrac{\zeta^2(-2n+1)}{B_{2n}^2}$

If the value of the series above is equal to $\dfrac{\pi^A}{B}$ , where $A$ and $B$ are integers, find $A+B$ .

Notations

$\zeta(\cdot)$ denote the Riemann zeta function .
$B_n$ denote the $n^\text{th}$ Bernoulli number.

The answer is 26.

1 solution

Aditya Kumar
Jan 14, 2016

We use the identity: $\zeta \left( -n \right) =\frac { -{ B }_{ n+1 } }{ n+1 } \\$

Therefore, on substituting -n as -2n+1, we get: $\frac { \zeta \left( -2n+1 \right) }{ { B }_{ 2n } } =\frac { -1 }{ 2n }$

Therefore, on taking summation we get, $\sum _{ n=1 }^{ \infty }{ { \left( \frac { \zeta \left( -2n+1 \right) }{ { B }_{ 2n } } \right) }^{ 2 } } =\sum _{ n=1 }^{ \infty }{ { \left( \frac { 1 }{ 2n } \right) }^{ 2 } } =\frac { \zeta \left( 2 \right) }{ 4 } =\frac { { \pi }^{ 2 } }{ 24 }$

+1. this is what i intended!

Aareyan Manzoor - 5 years, 5 months ago

A Squared Riemann Zeta

The answer is 26.

1 solution

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