Big-O: Why is the time complexity of these loops O(N)?

Question

I have the following function.

int fun(int n) 
{ 
  int count = 0; 
  for (int i = n; i > 0; i /= 2) 
     for (int j = 0; j < i; j++) 
        count += 1; 
  return count; 
} 

It has O(N) time complexity. I thought it's O(N LogN)(i.e. If we remove the inner loop we get)

for (int i = n; i > 0; i /= 2){
...
}

This is O(Log N) isn't it? And the inner loop turns out to be O(N) as it just increase linearly. So I think it's supposed to be O(Log N * N). What might go wrong on my thought?

Answer 1

Look at how many steps the inner loop is taking. It's $n + \frac{1}{2}n + \frac{1}{4}n + \cdots$. However the geometric series tells us:

$$\sum_{k=0}^{\lceil\log n\rceil +c}\frac{n}{2^k} < n\sum_{k=0}^\infty \frac{1}{2^k} = 2n$$