If you just need to show that the sum is O(n log n), you can show that
Σ log i ≤ Σ log n = n log n
Therefore, your function is O(n log n). If you want to be even more formal, you can use the constants c = 1 and n0 = 1.
The more interesting question is to show that the sum is Θ(n log n) by proving an Ω(n log n) lower bound. To do this, note that the sum is greater than or equal to the sum of the last n / 2 terms in the summation. Each of those terms in the summation is at least log (n / 2). This gives a lower bound of (n / 2) log(n / 2) = (n / 2) (log n - log 2), which is Ω(n log n). Therefore, your summation is O(n log n) and Ω(n log n), so it's Θ(n log n).
Hope this helps!
