[seqfan] Series of reciprocal harmonic numbers and its asymptotic

[Tomasz Ordowski]

Hello everyone again!

Integral dx / (log(x) + gamma) = exp(-gamma) Ei(log(x) + gamma) + C.
Integral dx / log(x) = li(x) + c, where li(x) = Ei(log x) as is well
known.

It's easy to prove that
lim_{n->oo} (H_n - n / Sum_{k=1..n} 1/H_k) = 1.
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
<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+>
Sum_{k=1..n} 1/H_k = n / (H_n - 1 + ...) =
 = n / (log(n) + gamma - 1 + O(1/log n)).
Cf. A096987 : https://oeis.org/A096987
See https://oeis.org/history/view?seq=A096987&v=100
That's a nice result, but let's move on ...

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

Is it provable that
lim_{n->oo} E(n) = +oo* and lim_{n->oo} L(n) = +oo** ?
Note that for small n, the value of L(n) is negative.
So it has to change sign to positive for large n
(beyond all computational possibilities).

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).

The last constant -E is hidden in a nice asymptotic of this sum:
Sum_{k=1..n} 1/H_k = exp(-gamma) (Ei(log(n) + gamma) - 1) + o(1).

Is it too good to be true?

Best,

Thomas
_______________________
(*) 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?
"Mathematica" gave up.
__________
(**) Similarly (click on this link):
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?
Notice that for small n this is negative [sic].
"Mathematica" hoisted the white flag.