Pi, Harmonic Numbers, Continued Fraction
Leroy Quet
qq-quet at mindspring.com
Tue Jun 22 19:38:48 CEST 2004
Here is a somewhat trite observation that, in any case, should be
interesting to some seq.fan readers.
I checked if the rational sequence's ({a(k)}'s) numerators or
denominators were in the EIS.
And, no, the numerators/denominators were not in the EIS.
Let {a(k)} be the sequence of positive and negative rationals, where each
a(m) is such that
the continued fraction
[a(1); a(2), a(3),..., a(m)]
is equal to
sum{k=1 to m} 1/k = H(m) =
the m_th harmonic number,
for all positive integers m.
So, {a(k)} begins:
1, 2, -5/4, 28/9, -81/64, 704/225, -325/256, 768/245,...
(Figured by hand, so maybe I erred.)
In any case, interestingly:
limit{m->oo} a(2m) = pi.
And
limit{m->oo} a(2m+1) = -4/pi.
Actually, this is seen from the closed form for the terms of the sequence:
a(1) =1, a(2) =2,
for m >= 3,
a(m) = (-1)^m (2m-1) ((m-2)!!)^2/((m-1)!!)^2,
where n!! is the "double factorial",
n!! = n*(n-2)*(n-4)*(n-6)...*(1 or 2).
I wonder, however,
what is the asymptotic nature of:
|H(2m) - [a(1); a(2), a(3),...,a(2m-1), pi]|
and
|H(2m+1) - [a(1); a(2), a(3),...,a(2m), -4/pi]|
as m -> oo ?
thanks,
Leroy Quet
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