Yes, that's right. Actually it would have been simpler to use the sum
1+1/3+...+1/(2n-1), i.e. with a minus in the denominator of the last term instead of a plus, so that the n tells how many terms in the sum. Then the integrand would have been simply (1-t^(2n))/(1-t^2). Forest ----- Original Message ----- From: Kristofer Munsterhjelm Date: Tuesday, May 25, 2010 12:42 pm > [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.) > ---- Election-Methods mailing list - see http://electorama.com/em for list info
