Massive Algorithms: A Time Complexity Question

void fun()
{
   int i, j;
   for (i=1; i<=n; i++)
      for (j=1; j<=log(i); j++)
         printf("GeeksforGeeks");
}

Time Complexity of the above function can be written as $\theta(\log 1) + \theta(\log 2) + \theta(\log 3) + . . . . + \theta(\log n)$ which is $\theta (\log n!)$

Order of growth of $\log n!$ and $n\log n$ is same for large values of $n$ , i.e., $\theta (\log n!) = \theta (n\log n)$ . So time complexity of fun() is $\theta (n\log n)$ .