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# Can anybody prove this about the zeta function?

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

Note by Ali Caglayan
2 years ago

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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. · 2 years 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. · 2 years ago

Very nice! · 2 years ago