Riemann $\zeta$ is back again

Calculus Level 3

Let $n$ be any natural number such that $\zeta(n)$ converges and $\mathbb{N} - 2\mathbb{N}$ be the set of all odd numbers. Then we can write

$\sum_{k \in (\mathbb{N} - 2\mathbb{N})} \frac{1}{k^n \zeta(n)} = x,$

for some real number $x$ . About the number $x,$ it is true that.

x^{-1}

is always even.

x \notin \mathbb{N}

For some

n, x > 1

For some

n, x \in \mathbb{N}

1 solution

Guilherme Niedu
Jan 7, 2019

$\large \displaystyle x = \frac{1}{\zeta(n) } \left [ \frac{1}{1^n} + \frac{1}{3^n} +\frac{1}{5^n} +... \right]$

$\large \displaystyle x = \frac{1}{\zeta(n) } \left [ \zeta(n) - \left (\frac{1}{2^n} + \frac{1}{4^n} +\frac{1}{6^n} +... \right ) \right]$

$\large \displaystyle x = \frac{1}{\zeta(n) } \left [ \zeta(n) - \frac{1}{2^n} \left (\frac{1}{1^n} + \frac{1}{2^n} +\frac{1}{3^n} +... \right ) \right]$

$\large \displaystyle x = \frac{1}{\zeta(n) } \left [ \zeta(n) - \frac{1}{2^n} \zeta(n) \right]$

$\color{#20A900} \boxed{ \large \displaystyle x = 1 - \frac{1}{2^n}}$

So, since $n \geq 2$ for $\zeta$ to converge, $0 < x < 1$ , or $\color{#3D99F6} \boxed { x \notin \mathbb{N} }$

Riemann ζ \zeta ζ is back again

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Riemann $\zeta$ is back again