Let me first define the problem:
Solution:
Mathematically we can prove that the sum of an Harmonic series is ln n + gamma( a constant). This constant is called Euler's constant and its approx value is 0.57721
Hence we can say F(n) = O(log n)
You can find the document here.
Prob: F(n) is a function defined as :
F(n) = 1 ; n = 1
F(n) = F(n-1) + 1/n ; n>1
Prove that Big Oh of F(n) is log n
Solution:
F(n) = F(n-1) + 1/n
= F(n-2) + 1/(n-1) + 1/n
=F(n-3) + 1/(n-2) + 1/(n-1) + 1/n
This kind of series is called the Harmonic Series and can be represented as 1 + 1/2 + 1/3 + 1/4 + ......+ 1/n
Mathematically we can prove that the sum of an Harmonic series is ln n + gamma( a constant). This constant is called Euler's constant and its approx value is 0.57721
Therefore, we can say limit f(n) = limit ln n + gamma
n->infinity n->infinity
Hence limit f(n)/log n
n->infinity
= limit (ln n + gamma)/log n
n->infinity
= limit ln n/log n + gamma/limit log n
n->infinity n->infinity
= ln 2 + 0
= ln 2
You can find the document here.