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I would like to do a toy verification of the Riemann hypothesis exploiting theLagarias theorem (see the section Applications in the following link) and the fact that we know a lot of decimals for Apèry's constant.

Combining Lagarias, with the Dirichlet series for the sum of divisor function $\sigma(n)$ and Euler series involving (as the last series involving $\pi$ in the following link) the nth harmonic number, $H_n$ we've

$$ \frac{\pi^2}{6} \left( \zeta(3)-\frac{\pi^2}{12} \right) < \sum_{n=1}^{\infty} \frac{e^{H_n}\log H_n}{n^3}, $$ that we can also write it as a infinite product taking exponentials $$e^{ \frac{\pi^2}{6} ( \zeta(3)-\frac{\pi^2}{12}) }<\lim_{N\to\infty} \prod_{n=1}^{N} \left( H_n \right) ^{e^{H_n}/n^3},$$ this is $$ \prod_{n=1}^{\infty} \left( H_n \right) ^{ \frac{e^{H_n}}{n^3}}.$$ My attempt to get bounds was use the more simplest inequalities involving harmonic numbers and logarithms (because this is one of the more hardest problems in mathematics should be virtually impossible use the Squeeze theorem). In previous link, that is Wikipedia's page for Harmonic Numbers one can find several asymptotic identities, expressions involving harmonic numbers under the sign integrals...

Since for $n>1$ we've that $e^{H_n}\log H_n=\int_1^{H_n}\frac{d}{dx} \left( e^x\log x \right)dx $ and in previous series $\sum_1^{\infty}=\sum_2^{\infty}$, RHS of previous inqualities (the first that we've stated) can be written as $$\sum_{n=2}^{\infty}\frac{1}{n^3}\int_1^{H_n}e^x\log xdx+\sum_{n=2}^{\infty}\frac{1}{n^3}\int_1^{H_n}\frac{e^x}{x}dx.$$

For the integral inside the first series Wolfram Alpha say that an antiderivative is $e^x\log x-Ei(x)$, where $Ei(x)$ is the exponential integral function, and the second is the same function $Ei$.

Question. Provide us a good approach to get a good approximation of RHS in $$ \frac{\pi^2}{6} \left( \zeta(3)-\frac{\pi^2}{12} \right) < \sum_{n=1}^{\infty} \frac{e^{H_n}\log H_n}{n^3}, $$ after you substract LHS, to study this difference. Thanks in advance.

My goal thus is see good computations with interesting formulas andasymptotic identities involving harmonic numbers, and see how you ensure your statements.

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  • $\begingroup$ Perhaps the paragraph concerning the computations with $Ei(x)$ has no mathematical sense, thus after there is an answer, to clean the post, I could delete it, always adding in the body of the post that I cut it. Thanks. $\endgroup$ – user243301 May 11 '16 at 9:41

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