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Sum of terms of continued fraction for n-th harmonic number, 1 + 1/2 + 1/3 + ... + 1/n.
9

%I #20 Mar 23 2017 04:30:15

%S 1,3,7,14,15,10,16,19,26,35,72,41,38,79,83,42,59,143,68,61,70,51,50,

%T 78,74,82,130,113,111,315,235,1190,211,407,112,122,142,246,693,133,

%U 138,162,1904,243,170,539,363,210,197,518,275,502,527,316,1729,224,228,909

%N Sum of terms of continued fraction for n-th harmonic number, 1 + 1/2 + 1/3 + ... + 1/n.

%C Is anything known about the asymptotics of this sequence?

%C Should be asymptotic to D*n^(3/2) with D=0.4.... - _Benoit Cloitre_, Dec 23 2003

%H G. C. Greubel, <a href="/A058027/b058027.txt">Table of n, a(n) for n = 1..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HarmonicNumber.html">Harmonic Number</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ContinuedFraction.html">Continued Fraction</a>

%e 1 + 1/2 +1/3 = 11/6 = 1 + 1/(1 + 1/5). So sum of terms of continued fraction is 1 + 1 + 5 = 7.

%t Table[Plus @@ ContinuedFraction[HarmonicNumber[n]], {n, 60}] (* _Ray Chandler_, Sep 17 2005 *)

%o (PARI) a(n) = vecsum(contfrac(sum(k=1, n, 1/k))); \\ _Michel Marcus_, Mar 23 2017

%Y m-th harmonic number H(m) = A001008(m)/A002805(m).

%Y Cf. A055573, A100398, A110020, A112286, A112287.

%K easy,nonn

%O 1,2

%A _Leroy Quet_, Nov 15 2000