For the purposes of this question, the Harmonic Number, $H_n$ is defined by the finite series sum $H_n=\sum_{k=1}^n\frac{1}{k}$ (n and k being positive integers) and the Harmonic Logarithm, $hlog(n)$ is here defined as $$hlog(n)=\left( H_{n^2}-H_{n} \right)=\sum_{k=n+1}^{n^2}\frac{1}{k}$$

The standard definition for $\gamma$ the Euler–Mascheroni constant is

$$\gamma=\lim_{n \rightarrow \infty} \gamma_n$$ where $\gamma_n=H_n-\log(n)$

Following Scott [REF2], let $a_n=2\gamma_n-\gamma_{n^2}$, giving $\gamma=\lim_{n \rightarrow \infty} a_n=\lim_{n \rightarrow \infty} \left(2\gamma_n-\gamma_{n^2} \right)$

$$\begin{align}a_n&=2\gamma_n-\gamma_{n^2}\\ &=2\left(H_n-\log(n) \right)-\left(H_{n^2}-\log(n^2) \right)\\ &=2H_n-2\log(n)-H_{n^2}-2\log(n)\\ &=2H_n-H_{n^2}\\ &=H_n-\left(H_{n^2}-H_n \right)\\ &=H_{n}-hlog(n)\\ \end{align}$$

Therefore in the $\lim_{n \rightarrow \infty}$ we have $$\gamma=\lim_{n \rightarrow \infty} \left( H_n-hlog(n)\right)$$ or coverging from below using the alternative definition $\gamma_n=H_n-\log(n+1)$ $$\gamma=\lim_{n \rightarrow \infty} \left( H_n-hlog(n+1)\right)\tag1$$

where $\gamma$ is the Euler–Mascheroni constant.

(1) gives a slowly converging rational series for $\gamma$ $$\gamma=\sum_{n=1}^{\infty} \frac{1}{n}+\frac{1}{n+1}-\sum_{k=n^2+1}^{(n+1)^2}\frac{1}{k}$$


For the theoretical proofs and background see for example:

[REF1] J. Lambek and L. Moser, (Feb., 1956), Rational Analogues of the Logarithm Function, The Mathematical Gazette, Vol. 40, No. 331, pp. 5-7.

[REF2] J. A. Scott, (Nov., 1996), The Euler Constant γ without Logarithms, , The Mathematical Gazette, Vol. 80, No. 489, pp. 585-586.

(Aside: The term "Harmonic Logarithm" appears to have been first coined by M.F. Egan in the same issue of the Mathematical Gazette as the Lambek and Moser paper, using a slightly different definition to mine above (see pages 8-10). If readers know of any other relevant papers please let me know. The term "Harmonic Logarithm" has more recently been applied to a definition involving both rational and transcendental numbers http://mathworld.wolfram.com/HarmonicLogarithm.html.)

Relation of the Harmonic Logarithm to the Prime Number Theorem

One way of stating the prime number theorem is

$$\pi(N) \thicksim \frac{N}{log(N)}$$

where $\pi(N)$ is the prime counting function, being the number of primes less than or equal to N.

A closer approximation to the prime counting function can be found using the $Li(x)=\int_2^x \frac{dt}{log t}$ function. However if we calculate $Li((N+1)^2)-Li((N)^2)$ we find that the number of primes between two consecutive squares, $N^2$ and $(N+1)^2$, is also approximately given by $\frac{N}{log(N)}$

Now one conjecture here involving series (2) is

$$\gamma_n \;(n^2)!= \left( \frac{1}{n}+\frac{1}{n+1}-\sum_{k=n^2+1}^{(n+1)^2}\frac{1}{k} \right) (n^2)! = \frac{C}{p_1 p_2 ... p_n}$$ where C is some arbitrary positive integer and $p_1 p_2 ... p_n$ is a product of all the primes between $n^2$ and $(n+1)^2$, none of which being divisors of C.

However perhaps of greater interest is how to tackle the proof of the prime number theorem involving $hlog(N)$ instead of $log(N)$ $$\pi(N) \thicksim \frac{N}{hlog(N)}$$

without involving transcendental functions in the proof at all.

I have seen proofs using $H_n$ that generate Chebyshev bounds for the prime counting function, as referenced in the J. Lambek and L. Moser paper above, but surely this can be improved upon using $hlog(N)$ instead.

  • $\begingroup$ Replacing $\text{Li}(x)$ by an approximation $f(x) = \text{Li}(x)+o(\text{Li}(x))$ won't change the proof of the prime number theorem at all. $\endgroup$ – reuns Aug 29 '17 at 12:35
  • $\begingroup$ In hindsight I need to make further changes to make this question clearer. The ultimate aim is to encourage more competent mathematicians than me to try to prove the prime number theorem without any recourse to transcendental functions or numbers. Therefore Li(x) cannot be used in the final proof. Something like $\sum_{k=2}^n \frac{1}{hlog(k)}$ seems to track Li(n) quite well, but I have no idea how to relate one to the other at the moment. $\endgroup$ – James Arathoon Aug 29 '17 at 13:13
  • $\begingroup$ As I said replacing $\text{Li}(x)$ by an approximation won't change anything. For the analytic proof of the PNT, $\text{Li}(x)$ is clearly the better choice because $\int_2^\infty (\pi(x)-\text{Li}(x)) x^{-s-1}dx$ is analytic at $s=1$ and $\int_2^\infty \text{Li}(x) x^{-s-1}dx$ is easy to understand (it is for good reasons if we write the PNT in term of $\text{Li}(x)$) $\endgroup$ – reuns Aug 29 '17 at 13:15

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