Crea sito
QUASI-HARMONIC SERIES


It's well known that the harmonic series Sum 1/n diverges, while the sum (from n=1 to infinity) Sum 1/n^s with s > 1 converges (and its value is the Riemann zeta function ζ(s).

It is therefore natural to consider the series in which the exponent depends on n,

Sum 1/n^(1+k(n))
with k(n) a decreasing function.

For k(n)=0 we have the harmonic series, that diverges. For k(n) constant and positive we have a coverging series. The interesting things occur when k(n) is positive, and approaches 0 as n goes to infinity. For example k(n)=1/n, or k(n)=1/ln(n), or k(n)=1/sqrt(ln(n)). In the first two cases the sum is still diverging. In the last one it converges.

Is there a function the rate of decrease of which is the borderline between convergence and divergence ? In other words, is there a function ko(n) such that if limn -> inf k(n)/ko(n) = 0 than the sum diverges, and if the limit is infinity it converges ?

For k(n) = (ln(ln(n)))-m (m>=0) the sum converges.
For k(n) = (ln(ln(n)))- ln(ln(n)) the sum diverges.
For k(n) = (ln(ln(n)))- ln(ln(ln(n))) = exp( - (ln(ln(ln(n))))2 ) the sum converges.

Can you pin down better the borderline of divergence ?

Marco Corvi - 2013