# A series whose convergence is equivalent to the Riemann hypothesis

It was claimed here that the convergence of the series$$\sum_{n=2}^\infty \frac{\Lambda(n)-1}{n^{1/2}\log^3 n}\tag1$$(where $$\Lambda$$ is the Von Mangoldt function) is equivalent to the Riemann hypothesis. Is this true? That post provided a link to the Wikipedia article about the Von Mangoldt function, which does not mention this. Also, this page about the Von Mangoldt function in the context of the Riemann hypothesis makes no mention to that.

If it is true that the convergence of the series $$(1)$$ is equivalent to the Riemann hypothesis, then I would like to have a reference for that.

Claymath official description of the Riemann hypothesis claims the RH is true iff $$\pi(x) = Li(x)+O(x^{1/2}\log x)$$ so $$\psi(x) = x+O(x^{1/2}\log^2 x)$$ and I have been quite sloppy in that it implies the RH is true iff $$\sum_{n=2}^\infty \frac{\Lambda(n)-1}{n^{1/2}\log^{\color{red}{3+\epsilon}} n} < \infty\tag{1}$$

as with partial summation $$\sum_{n \le x} (\Lambda(n)-1) \frac{1}{n^{1/2}\log^{a} n}=\frac{\psi(x)-x}{x^{1/2}\log^a x}+\sum_{n \le x-1} (\psi(n)-n)O(\frac{1}{n^{3/2}\log^a n})\\ = O(\log^{2-a}(x))+\sum_{n \le x} O(\frac{1}{n\log^{a-2} n})$$

The point is to show an effective explicit formula (p.28) $$\psi(x) =\sum_{n \le x} \Lambda(n)= x - \sum_{|\Im(\rho)| \le T} \frac{x^{\rho}}{\rho}+O(\frac{x\log^2 x}{T})=x - \sum_{k\le K} 2\Re(\frac{x^{\rho_k}}{\rho_k})+O(\frac{x\log^2 x}{K/\log K})$$ where $$K = N(T)$$ and the density of zeros gives $$K \sim C T \log T,T \sim c K/\log K$$,$$\Im(\rho_k) \sim c k/\log k$$.

In this form, under the RH, with $$K = x^{1/2}$$ it yields $$\psi(x) =x +O(x^{1/2}\log^{2+\delta})$$

Plotting those things indicates the series may converge very slowly with $$\epsilon = 0$$ and it is quite certain (under the RH) it converges with $$\epsilon = 2$$.

A good place to start is by reading Terry Tao's post "The Riemann hypothesis in various settings"

• so (if we can't find a cancellation in the phases of the explicit formula) we really need $\log^{3+\epsilon} n$ instead of $\log^3 n$ Commented Jan 21, 2019 at 18:25