[seqfan] Sums of harmonic errors

[Tomasz Ordowski]

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!