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You can read Lagarias equivalence to the Riemann's Hypothesis from this MathWorld, I am saying (5), or well from the reference [1].

Let for real numbers $0<x<1$ the factor $x^n$, where $n\geq 1$ is an integer. Then we multiply Lagarias statement by this factor and after we take the sum from $n=1$ to $\infty$. Since the Lambert series for the sum of divisor function is $\sum_{n=1}^\infty\frac{nx^n}{1-x^n}$, and a generating function for harmonic numbers is $-\frac{\log(1-x)}{1-x}$ we conclude that on assumption of the Riemann's Hypothesis $$\sum_{n=1}^\infty\frac{nx^n}{1-x^n}\leq\frac{\log(1-x)}{x-1}+\sum_{n=1}^\infty\left(e^{H_n}\log\left(H_n\right)\right)x^n.$$

You can find these generating functions for example in the corresponding Wikipedia's entries for Lambert series and Harmonic number.

Question. I believe that should be very difficult to calculate this generating function $$g(x):=\sum_{n=1}^\infty\left(e^{H_n}\log\left(H_n\right)\right)x^n$$ for $0<x<1$. But if our purpose is explore if it is possible to calculate it, what calculations can be done? That is, that I am asking, since my belief is that calculate the generating function is very difficult, is what could be the first calculations to try get it or well get a relevant property of such generating function? Summarizing, imagine that we need to study such generating function, how to start analyzing it? Thus I am asking about the first ideas, reasonings or hints. Thanks in advance.

Reference:

[1] Lagarias, An Elementary Problem Equivalent to the Riemann Hypothesis, The American Mathematical Monthly, Vol. 109, No. 6 (2002).

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  • $\begingroup$ Thus that I am asking is what strategy and calculations can you explain me with the purpose to study the generating function $\sum_{n=1}^\infty\left(e^{H_n}\log\left(H_n\right)\right)x^n$, where $H_n$ is the $nth$ harmonic number. What are the first calculations, even if we can not get in a closed-form $g(x)$, that we need to do? Many thanks all users. $\endgroup$ – user243301 Apr 26 '17 at 20:29
  • $\begingroup$ Since there was a downvote, if this is a bad question because aren't feasible good calculations please add a comment, or explain you reasonings to know it. Many thanks. $\endgroup$ – user243301 Apr 26 '17 at 21:54
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    $\begingroup$ The best RH equivalent to work with is historically the first, due to Nicolas. In French with English abstract, math.univ-lyon1.fr/~nicolas/petitsphiDPP.pdf See also arxiv.org/abs/1012.3613 I am pretty sure i have posted numerical evidence for this in some MSE answers. Much, much easier to work with than Superior Highly Composite or Colossally Abundant numbers $\endgroup$ – Will Jagy Apr 26 '17 at 23:44
  • $\begingroup$ Many thanks for your remarks @WillJagy $\endgroup$ – user243301 Apr 27 '17 at 8:42
  • $\begingroup$ With this BOUNTY, then I hope encourage to study this function $$g(x):=\sum_{n=1}^\infty\left(e^{H_n}\log\left(H_n\right)\right)x^n,$$ I don't know if is well defined for all $0<x<1$. I am waiting answers showing remarkable properties of this $g(x)$ even if a closed-form isn't possible. Many thanks all users. $\endgroup$ – user243301 May 6 '17 at 22:14

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