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
·
2 years, 3 months ago

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@Michael Mendrin
–
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
·
2 years, 3 months ago

@Michael Mendrin
–
Interesting method. I tried squeezing it however I am having trouble establishing a good integer upper bound.
–
Ali Caglayan
·
2 years, 3 months ago

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TopNewestFirst, 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 · 2 years, 3 months ago

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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 · 2 years, 3 months ago

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– Ali Caglayan · 2 years, 3 months ago

Very nice!Log in to reply

– Ali Caglayan · 2 years, 3 months ago

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