# Salvaging proofs

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.

Note by Jake Lai
3 years, 5 months ago

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Since 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, $2{ \zeta (s) }^{ 2 }<2\zeta (s+1)\zeta (s-1)<{ \zeta (s+1) }^{ 2 }+{ \zeta (s-1) }^{ 2 }$ $4{ \zeta (s) }^{ 2 }<{ \zeta (s+1) }^{ 2 }+{ \zeta (s-1) }^{ 2 }+2\zeta (s+1)\zeta (s-1)$ $2\zeta (s)<{ \zeta (s+1) }+{ \zeta (s-1) }$

- 3 years, 5 months ago

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.

- 3 years, 5 months ago

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

- 3 years, 5 months ago

Same thing, hehe.

- 3 years, 5 months ago

Maybe I misunderstood the notation...

- 3 years, 5 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.

- 3 years, 5 months ago

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

- 3 years, 5 months ago

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

- 3 years, 3 months ago