Riemann ζ \zeta is limitless

Calculus Level 4

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

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

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

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


The answer is 1.

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1 solution

Guilherme Niedu
Jan 6, 2019

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

x ( n ) = 1 ζ ( n ) [ ζ ( n ) ( 1 2 n + 1 4 n + 1 6 n + . . . ) ] \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]

x ( n ) = 1 ζ ( n ) [ ζ ( n ) 1 2 n ( 1 1 n + 1 2 n + 1 3 n + . . . ) ] \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]

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

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

So:

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

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