This follows easily from a simple fact in Calculus:
and we have the following inequality:
Here we can conclude that S = 1 + 1/2 + … + 1/n is both Ω(log(n)) and O(log(n)), thus it is Ɵ(log(n)), the bound is actually tight.
This follows easily from a simple fact in Calculus:
and we have the following inequality:
Here we can conclude that S = 1 + 1/2 + … + 1/n is both Ω(log(n)) and O(log(n)), thus it is Ɵ(log(n)), the bound is actually tight.