A096987 Numerator of Sum_{k=1..n} 1/H(k), where H(k) = Sum_{j=1..k} 1/j is the k-th harmonic number.

0, 1, 5, 73, 2221, 353777, 19595573, 239046803, 198972350083, 1535302297058707, 100536661265514127, 8974880059175708288297, 818810519369821323965929237, 990666575600755815615137883006341, 1220749860499992165560973207703210595953

Offset: 0

Author

Leroy Quet, Aug 19 2004

Keywords

• easy
• frac
• nonn

Examples

			1/1 + 1/(1 + 1/2) + 1/(1 + 1/2 + 1/3) = 73/33, so a(3) = 73.
		

Crossrefs

Cf. A124432 (denominators), A000720, A001008, A002805, A001620, A080130.

Programs

Mathematica

f[n_] := Numerator[ Sum[ 1/HarmonicNumber[j], {j, 1, n}]]; Table[ f[n], {n, 0, 14}] (* Robert G. Wilson v, Aug 21 2004 *)

PARI

m=13;for(n=0,m,print1(numerator(sum(k=1,n,1/sum(j=1,k,1/j))),",")) \\ Klaus Brockhaus, Aug 21 2004

Formula

From Thomas Ordowski, Mar 21 2023: (Start)

Sum_{k=1..n} 1/H(k) ~ Sum_{k=2..n} 1/log(k) ~ Integral_{2..n} dx/log(x) = Li(n).

Sum_{k=1..n} 1/H(k) = Sum_{k=1..n} 1/(log(k + 1/2) + gamma) - C + o(1), where gamma = A001620 = 0.577... is Euler's constant and the constant C = 0.0229825...

Sum_{k=1..n} 1/H(k) = exp(-gamma)*(Ei(log(n) + gamma) - 1) + o(1), where Ei(x) is the exponential integral function of real x, and we have Ei(log(x)) = li(x).

Note that a(n)/A124432(n) ~ pi(n) = A000720(n), see my first formula.

Sum_{k=1..n} 1/H(k) = n/(H(n) - 1 + ...) = n/(log(n) + gamma - 1 + O(1/log(n))).

Theorem: lim_{n->oo} (H(n) - n / Sum_{k=1..n} 1/H(k)) = 1, see my third formula.

Proof: since Integral dx / (log(x) + gamma) = exp(-gamma)*Ei(log(x) + gamma) + c, so we get lim_{n->oo} (log(n) + gamma - n*exp(gamma) / Ei(log(n) + gamma)) = 1, qed. (End)

Extensions

More terms from Klaus Brockhaus and Robert G. Wilson v, Aug 21 2004