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

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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