Does my solution converge to O(N) for worst-case time complexity? Forgive me if this should be in StackOverflow or Mathematics instead!

I was given the following question at an interview:

Given an array of unique integers, find the first missing non-negative integer that is missing to form a consequence sequence of non-negative integers. 

For example, the input [3, 4, -1, 0, 1] should give 2. 
The input [1, -3, 2, 0] should give -1 (i.e no missing numbers).
The input of [-2, -3, -5] should give 0


I came up with a solution that is something akin to here: https://play.golang.org/p/b2pXr9kZYxM:

func findMissingNumber(nums []int) int{ 
    j := 0
    for i, num := range nums {
        if num &lt; 0 {
            nums[i], nums[j] = nums[j], nums[i]
            j++
        }
    }   
    if j &gt;= len(nums) {
        return 0
    }
    count := 0
    for i:=j ;i &lt; len(nums);{
        count++ 
        if nums[i] == i-j {
            i++
            continue
        }
        if nums[i]+j &gt;= len(nums) || nums[nums[i]+j]+j &gt;= len(nums) {
            nums[i] = -1
            i++
            continue
        }
        temp := nums[nums[i]+j]
        nums[nums[i]+j]= nums[i]
        nums[i] = temp  
    }
    fmt.Println(count)
    for i := j; i &lt; len(nums); i++ {
        if nums[i] == -1 {
            return i - j
        }
    }
    return -1
}


Both the interviewer and I agreed that my solution could've been slightly more optimized (GeekForGeek seems to agree) but we were unsure if the worst-case for my solution was $O(N)$ or $O(N^2)$. 

It seems like my solution has $$ O(N) + O(N/2) + O(N/4) + O(N/6) +\cdots+ O(1) $$ which becomes essentially:
$$\frac{N}{2}\sum_{n=1}^{N} n^{-1}$$

My calculus is a bit rusty (as are my algorithm skills) but I know this is a series diverges for $N \to \infty$. But can I say that for N sufficiently smaller than infinite (i.e Integer.Max), my worst-case becomes simply $$O(kN) = O(N)?$$
$ \sum_{i=1}^n \frac{i}{n} $ is the harmonic sequence, which goes to $\log(n)$

since in your question you claim the $T =\frac{n}{2} \sum_{i=1}^n \frac{i}{n}$ 

$T = \frac{n}{2} \log(n) = O(n\log(n))$