[email protected] wrote:
Note that the sum
1+1/3+…+1/(2n+1)
is the integral (with respect to t) from zero to one of the sum
1+t^2+…+t^(2n),
and that this integrand is a finite geometric sum with closed form
(1-t^(2n+1))/(1-t^2) .
So this is the appropriate integrand for a Sainte-Lague version that
allows fractional values of n, i.e. that works with any kind of range
ballot, not just approval.
For the list, I'll note that the 2n+1 in the integrand appears to really
be 2(n+1).
This gives
f(x) = H(x + 0.5)/2 + ln(2),
where H is the harmonic number function, so an approximation to H(x) can
be used here as well.
(If the term had been 2n+1, then we'd have got f(x) = H(x)/2 + ln(2),
which is not right.)
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