Can anybody prove this about the zeta function?

Does the following always hold? And if so can you prove it $\left\lfloor\sum_{k=2}^n\zeta(k)\right\rfloor=n-1$

#RiemannZetaFunction #Summation #FloorFunction

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

First, observe that

$\displaystyle\sum _{ k=2 }^{ \infty }{ \left( \zeta \left( k \right) -1 \right) } =\displaystyle\sum _{ k=2 }^{ \infty }{ \left( \displaystyle\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ k } } } -1 \right) } =\displaystyle\sum _{ n=2 }^{ \infty }{ \sum _{ k=2 }^{ \infty }{ \frac { 1 }{ { n }^{ k } } } } =1$

In other words, the sum of all the fractions of $\zeta \left( n \right)$ from $n=2$ to infinity is exactly 1. With a bit of handwaving here, it can be surmised that the sum of $n$ Zeta functions starting at $n=2$ is $n-1+a$ , where $0<a<$ . Hence the floor value is $n-1$ .

This proof-toid is incomplete because we need to prove additional lemmas, such as $\zeta \left( n \right) >\zeta \left( n+1 \right) >1$ for all $n$ . But this is the gist of the proof anyway.

Michael Mendrin - 6 years, 8 months ago

Indeed, $\sum_{k = 2}^\infty (\zeta(k) - 1) = 1$ is the key identity. The rest of the proof is quite simple.

Each term $\zeta(k) - 1$ is positive. It follows that $0 < \sum_{k = 2}^n (\zeta(k) - 1) < 1.$ Then $n - 1 < \sum_{k = 2}^n \zeta(k) < n,$ as desired.

Jon Haussmann - 6 years, 8 months ago

Very nice!

A Former Brilliant Member - 6 years, 8 months ago

Interesting method. I tried squeezing it however I am having trouble establishing a good integer upper bound.

A Former Brilliant Member - 6 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$