The easiest way is through this piece of logic: while a number can be prime and greater than 1, try dividing it by prime numbers greater than its square root. For example, say we have 6-- it's square root is ~2.449. So we try to divide it by 3, the next greater prime number, but we realize it is divisible and thus not prime. We'll expand on the logic for this shortly!

1for i in range(2, n+1):
2    isPrime = True
3    for j in range(2, int(math.sqrt(i))+1):
4        if i % j == 0:
5            isPrime = False
6            break
7    if isPrime:
8        res += i

Complexity of Final Solution

Let n be the input number. We fill an array of size n to keep track of potential prime candidates for O(n) space complexity, and our while loop that updates our array performs roughly log n work over n iterations for O(n log n) time complexity.