Asymptotic complexity of T(n)=T(n-1)+1/n [closed]

Answer 1

For more details:

With H(N) = 1 + 1/2 + 1/3 + ... + 1/N

the function x :-> 1/x is a decreasing function so :

We sum from 1 to N the left part and for the right part we sum from 2 to N and we add 1, we get:

Then we calculate the left and right parts : ln(N+1) <= H(N) <= 1 + ln(N)

this implies H(N)/ln(N) -> 1 hence H(N)=Θ(log(N))

(from http://fr.wikipedia.org/wiki/S%C3%A9rie_harmonique#.C3.89quivalent_de_Hn)

Answer 2

It can be easily seen (or proven formally with induction) that T(n) is the sum of 1/k for the values of k from 1 to n. This is the nth harmonic number, H_n = 1 + 1/2 + 1/3 + ... + 1/n.

Asymptotically, the harmonic numbers grow on the order of log(n). This is because the sum is close in value to the integral of 1/x from 1 to n, which is equal to the natural logarithm of n. In fact, H_n = ln(n) + γ + O(1/n) where γ is a constant. From this, it is easy to show that T(n) = Θ(log(n)).