[seqfan] Re: Harmonic Constant
Tomasz Ordowski
tomaszordowski at gmail.com
Fri Mar 31 09:06:00 CEST 2023
ACKNOWLEDGMENTS:
thanks to all for private correspondence [off-list] on this matter.
The last formula in my draft to A096987
https://oeis.org/history/view?seq=A096987&v=43
completes the proof of my conjecture (in comments)
that lim_{n->oo} ( H(n) - n / Sum_{k=1..n} 1/H(k) ) = 1,
where H(m) = 1+1/2+...+1/m is the harmonic number.
Let's rewrite this formula in a slightly more precise notation:
Sum_{k=1..n} 1/H(k) = exp(-gamma) Ei(log(n)+gamma) - E + o(1).
Hence we have the following definition of the new constant
E = lim_{n->oo} (exp(-gamma) Ei(log(n)+gamma) - Sum_{k=1..n} 1/H(k)).
Note: the possible divergence of this limit does not refute my conjecture.
What is the approximate value of this constant (if it exists) ?
Maybe someone will try to calculate it.
T. Ordowski
czw., 30 mar 2023 o 15:27 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):
> 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):
>
>> Dear readers!
>>
>> 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!
>>
>>
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