I botched up a proof, but I salvaged useful bits and bobs from it.

We prove \(\zeta^2(s) < \zeta(s+1)\zeta(s-1)\) by recalling the Dirichlet series of \(\phi\), the Euler totient function, that is,

\[\sum_{n=1}^{\infty} \frac{\phi(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)}\]

Now we know that, for all \(n \geq 2\),

\[\frac{\phi(n)}{n^{s+1}} < \frac{\phi(n)}{n^s} \longrightarrow \sum_{n=2}^{\infty} \frac{\phi(n)}{n^{s+1}} < \sum_{n=2}^{\infty} \frac{\phi(n)}{n^s} \quad \therefore \frac{\zeta(s)}{\zeta(s-1)} < \frac{\zeta(s+1)}{\zeta(s)}\]

\[\therefore \zeta^2(s) < \zeta(s+1)\zeta(s-1)\]

Personally, I think it is interesting, as it invokes a Dirichlet series in a somewhat unexpected way. Also, this implies that \(\lbrace \frac{\zeta(n+1)}{\zeta(n)} \rbrace _{n=2}\) is monotonically increasing.

Challenge: Prove \(2\zeta(s) < \zeta(s+1)+\zeta(s-1)\), using the above result or by the trivial inequality.

## Comments

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TopNewestSince if \(a,b>0\), \(a^{2}+b^{2}>2ab\)

\[2\zeta (s+1)\zeta (s-1)<{ \zeta (s+1) }^{ 2 }+{ \zeta (s-1) }^{ 2 }\]

Following the note's inequality, \[2{ \zeta (s) }^{ 2 }<2\zeta (s+1)\zeta (s-1)<{ \zeta (s+1) }^{ 2 }+{ \zeta (s-1) }^{ 2 }\] \[4{ \zeta (s) }^{ 2 }<{ \zeta (s+1) }^{ 2 }+{ \zeta (s-1) }^{ 2 }+2\zeta (s+1)\zeta (s-1)\] \[2\zeta (s)<{ \zeta (s+1) }+{ \zeta (s-1) }\] – Julian Poon · 11 months ago

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There is also a proof without the result of the challenge by the trivial inequality, which is kind of obvious. – Jake Lai · 11 months ago

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Just prove that \( \zeta(s) \) concave downwards, the inequality follows. – Pi Han Goh · 11 months ago

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– Jake Lai · 8 months, 4 weeks ago

I just realised, Cauchy-Schwarz on \(\ell^2\) works.Log in to reply

I think you meant \(\lbrace \frac{\zeta(n+1)}{\zeta(n)} \rbrace _{n\ge2}\) – Julian Poon · 11 months ago

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– Jake Lai · 11 months ago

Same thing, hehe.Log in to reply

– Julian Poon · 11 months ago

Maybe I misunderstood the notation...Log in to reply

– Jake Lai · 11 months ago

Yours and mine both mean the same thing. I've seen both used before, and they all mean what we're both trying to say.Log in to reply