Riemann $\zeta$ is limitless

Calculus Level 4

The present problem is a continuation of the problem Riemann $\zeta$ is back again . Now that you've explored some of the properties of the function $x: S \subseteq \mathbb{N} \rightarrow \mathbb{R},$ where $S$ is the set of natural numbers such that $\zeta(n)$ converges, defined as

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

where $\mathbb{N} - 2\mathbb{N}$ is the set of all odd numbers, we would like to know how this function behaves as $n \rightarrow \infty,$ i.e., compute

$\lim_{n \rightarrow \infty} x(n).$

The answer is 1.

1 solution

Guilherme Niedu
Jan 6, 2019

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

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

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

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

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

So:

$\color{#3D99F6} \boxed{ \large \displaystyle \lim_{n \rightarrow \infty} x(n) = 1 }$

Riemann ζ \zeta ζ is limitless

The answer is 1.

1 solution

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Riemann $\zeta$ is limitless