# [seqfan] Re: Harmonic Constant

Tomasz Ordowski tomaszordowski at gmail.com
Thu Mar 30 15:27:03 CEST 2023

```P.S. Regarding my conjecture:

lim_{n->oo} ( H(n) - n / Sum_{k=1..n} 1/H(k) ) = 1.

Proof:

https://www.wolframalpha.com/input?i=lim+%28log%28n%29+%2B+gamma+-+n+e%5Egamma%2FEi%28log%28n%29+%2B+gamma%29%29+as+n-%3Eoo+

Note that  Integral dx / (log(n) + gamma) = exp(-gamma) Ei(log(x) +
gamma) + c.

In the special case, without the gamma constant, we have

Ei(log x) = li(x) and lim_{x->oo} ( log(x) - x / li(x) ) = 1.

Cf. the initial formula (below).

T. Ordowski

sob., 25 mar 2023 o 13:19 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):

>
> As is well known, log(x) - x/Li(x) ~ 1.
> It has also been proven that  log(x) - x/pi(x) ~ 1.
> Cf. https://en.wikipedia.org/wiki/Legendre%27s_constant
> Note that Sum_{k=1..n} 1/H(k) ~ Sum_{k=2..n} 1/log(k) ~
> ~ Integral_{2..n} dx/log(x) = li(n) - li(2) = Li(n) ~ pi(n).
> So we have  A096987(n) / A124432(n) ~ pi(n).
> See my draft (new comment and my formula):
> https://oeis.org/history/view?seq=A096987&v=29
>
> Let's finally define this Harmonic Constant:
> H = lim_{n->oo} (log(n) - n / Sum_{k=1..n} 1/H(k)),
> where the harmonic number H(k) = 1+1/2+...+1/k.
>
> Conjecture:
> the value of this constant H = 1 - gamma,
> where gamma = 0.577... is Euler's constant.
>
> Equivalently:
> lim_{n->oo} (H(n) - n / Sum_{k=1..n} 1/H(k)) = 1.
> Nice!
>
> Is this a known result or provable?
> Numerical verification will be hard.*
>
> Best regards,
>
> Thomas Ordowski
> _________________________
> (*) Calculations up to n = 10^15 should give the desired result,
> as suggested by a numeric sample from Amiram Eldar. Thanks!
>
>
```