Salvaging proofs

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

We prove ζ2(s)<ζ(s+1)ζ(s1)\zeta^2(s) < \zeta(s+1)\zeta(s-1) by recalling the Dirichlet series of ϕ\phi, the Euler totient function, that is,

n=1ϕ(n)ns=ζ(s1)ζ(s)\sum_{n=1}^{\infty} \frac{\phi(n)}{n^s} = \frac{\zeta(s-1)}{\zeta(s)}

Now we know that, for all n2n \geq 2,

ϕ(n)ns+1<ϕ(n)nsn=2ϕ(n)ns+1<n=2ϕ(n)nsζ(s)ζ(s1)<ζ(s+1)ζ(s)\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)}

ζ2(s)<ζ(s+1)ζ(s1)\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 {ζ(n+1)ζ(n)}n=2\lbrace \frac{\zeta(n+1)}{\zeta(n)} \rbrace _{n=2} is monotonically increasing.

Challenge: Prove 2ζ(s)<ζ(s+1)+ζ(s1)2\zeta(s) < \zeta(s+1)+\zeta(s-1), using the above result or by the trivial inequality.

Note by Jake Lai
3 years, 11 months ago

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Since if a,b>0a,b>0, a2+b2>2aba^{2}+b^{2}>2ab

2ζ(s+1)ζ(s1)<ζ(s+1)2+ζ(s1)22\zeta (s+1)\zeta (s-1)<{ \zeta (s+1) }^{ 2 }+{ \zeta (s-1) }^{ 2 }

Following the note's inequality, 2ζ(s)2<2ζ(s+1)ζ(s1)<ζ(s+1)2+ζ(s1)22{ \zeta (s) }^{ 2 }<2\zeta (s+1)\zeta (s-1)<{ \zeta (s+1) }^{ 2 }+{ \zeta (s-1) }^{ 2 } 4ζ(s)2<ζ(s+1)2+ζ(s1)2+2ζ(s+1)ζ(s1)4{ \zeta (s) }^{ 2 }<{ \zeta (s+1) }^{ 2 }+{ \zeta (s-1) }^{ 2 }+2\zeta (s+1)\zeta (s-1) 2ζ(s)<ζ(s+1)+ζ(s1)2\zeta (s)<{ \zeta (s+1) }+{ \zeta (s-1) }

Julian Poon - 3 years, 11 months ago

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Or that. I think that spotting that there's a way to slip in AM-GM is easier, but again, that just depends on what prior experience one has.

There is also a proof without the result of the challenge by the trivial inequality, which is kind of obvious.

Jake Lai - 3 years, 11 months ago

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I think you meant {ζ(n+1)ζ(n)}n2\lbrace \frac{\zeta(n+1)}{\zeta(n)} \rbrace _{n\ge2}

Julian Poon - 3 years, 11 months ago

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Same thing, hehe.

Jake Lai - 3 years, 11 months ago

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Maybe I misunderstood the notation...

Julian Poon - 3 years, 11 months ago

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@Julian Poon 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.

Jake Lai - 3 years, 11 months ago

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Just prove that ζ(s) \zeta(s) concave downwards, the inequality follows.

Pi Han Goh - 3 years, 11 months ago

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I just realised, Cauchy-Schwarz on 2\ell^2 works.

Jake Lai - 3 years, 9 months ago

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