[seqfan] Re: Sums of harmonic errors

[Tomasz Ordowski]

E(n) = exp(-gamma) Ei(log(n) + gamma) - Sum_{k=1..n} 1/H_k.
L(n) = li(n) - Sum_{k=1..n} 1/(H_k - gamma).

E = lim_{n->oo} E(n) is a constant
if and only if
L = lim_{n->oo} L(n) is a constant.

If so, these constants can be related by a simple expression:

E = lim_{n->oo} E(n) = exp(L), where L = lim_{n->oo} L(n).

Conjecture: L= -gamma and E = exp(-gamma).

Is it too good to be true?


niedz., 16 kwi 2023 o 17:07 Tomasz Ordowski <tomaszordowski at gmail.com>
napisał(a):

> AT THE END.
>
> Contrary to my conjecture:
>
> https://www.wolframalpha.com/input?i=Limit%5BExp%5B-EulerGamma%5D*ExpIntegralEi%5BLog%5Bn%5D+%2B+EulerGamma%5D+-+Sum%5B1%2FHarmonicNumber%5Bk%5D%2C+%7Bk%2C+1%2C+n%7D%5D%2C+%7Bn+-%3E+Infinity%7D%5D
> But is this result +oo reliable?
>
> Similar limit, but looking more familiar:
> lim_{n->oo} (li(n) - Sum_{k=1..n} 1/(H_k - gamma)) = ?
>
> https://www.wolframalpha.com/input?i=Limit%5BLogIntegralLi%5Bn%5D+-+Sum%5B1%2F%5BHarmonicNumber%5Bk%5D+-+EulerGamma%5D%2C+%7Bk%2C+1%2C+n%7D%5D%2C+%7Bn+-%3E+Infinity%7D%5D
> This page returns +oo again.
> If so, for what n does this difference change sign from negative to
> positive?
> Notice that for small n this is negative.
>
> Best,
>
> Tom ORDO
>
> czw., 13 kwi 2023 o 05:34 Tomasz Ordowski <tomaszordowski at gmail.com>
> napisał(a):
>
>> P.S. Conjecture:
>> E = exp(-gamma) = A080130 = 0.56145948356688516982414321479...
>> Sum_{k=1..n} 1/H_k = exp(-gamma) (Ei(log(n) + gamma) - 1) + o(1).
>> Cf. A096987 : https://oeis.org/history/view?seq=A096987&v=98
>>
>> T. Ordowski
>> ___________________
>> https://oeis.org/A080130
>> https://oeis.org/A096987
>>
>> śr., 5 kwi 2023 o 18:42 Tomasz Ordowski <tomaszordowski at gmail.com>
>> napisał(a):
>>
>>> Hello everyone again!
>>>
>>> Alabdulmohsin (2018) derived closed-form expression for the sum [sic]:
>>> Sum_{n >= 1} (log(n) + 1/(2n) + gamma - H_n) = (log(2 pi) - 1 -
>>> gamma)/2.
>>> Cf.
>>> https://en.wikipedia.org/wiki/Euler%27s_constant#Asymptotic_expansions
>>> The exact value of a similar sum is easier to determine, namely:
>>> Sum_{n >= 1} (H_n - log(n+1/2) - gamma) = (2 gamma + 1 - log(8))/2.
>>> With estimate: 1/(24(n+1)^2) < H_n - log(n+1/2) - gamma < 1/(24n^2).
>>> Cf. https://en.wikipedia.org/wiki/Euler%27s_constant#Series_expansions
>>>
>>> I also calculated approximate values of other sums:
>>> Sum_{n >= 1} (1/H_n - 1/(log(n) + 1/(2n) + gamma)) = 0.0867629...
>>> C = Sum_{n >= 1} (1/(log(n+1/2) + gamma) - 1/H_n) = 0.0229825...
>>> Problem: are there closed-form expressions for these constants?
>>> Cf. A096987 (see the second formula):
>>> https://oeis.org/A096987
>>> See again A096987 (the third formula).
>>> Note that Integral dx / (log(x) + gamma) =
>>> = exp(-gamma) Ei(log(x) + gamma) + c,
>>> where Ei(x) is the exponential integral function of real x,
>>> and as is well known, Ei(log x) = li(x), which looks familiar.
>>> Let's rewrite my third 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 to compute of my new constant
>>> E = lim_{n->oo} (exp(-gamma) Ei(log(n) + gamma) - Sum_{k=1..n} 1/H_k).
>>> Hard task: find a good numerical approximation of this constant (if it
>>> exists).
>>>
>>> Best,
>>>
>>> Thomas
>>> _________________
>>> See also my theorem with a short proof
>>> at the end of my new entry to A096987:
>>> https://oeis.org/A096987 (numerators) /
>>> https://oeis.org/A124432 (denominators),
>>> where is also my new comment. Thanks!
>>>
>>>