Asymptotic runtime of a nested sum i have to calculate the asymptotic runtime of a nested sum:

Sum $\sum_{n=1}^{log_{4}(n)} \sum_{n=1}^{\sqrt{n}}1$

My solution is $\Theta(log(n) \cdot \sqrt{n})$, which is wrong, because the solution says $\Theta(\sqrt{n})$. I dont understand where my mistake is:

Sum $\sum_{i=1}^{log_{4}(n)} \sum_{j=1}^{\sqrt{n}}1 =  \sum_{i=1}^{log_{4}(n)} \sqrt{n} = \sqrt{n} \sum_{i=1}^{log_{4}(n)}1 = \sqrt{n} \cdot (log(n)+1) =$ 
 $ log(n) \cdot \sqrt{n} + \sqrt{n} \stackrel{?}{=} \Theta(log(n) \cdot \sqrt{n}) $ 

Maybe my formula is wrong. Here is the pseudo-code i have to analyse:

someFunc(n)
  i = n
  while (i &gt;= 1)
   i=i/4;
   //we do sqrt(i) operations
   for j=1 to sqrt(i)
     do_some_atomic_op() //O(1)
   end
   //i will reduce in every step!
  end
end
In fact you do change the value of $n$, see $n=n/4$, and so the while-loop results in  $\sum_{i=1}^{\log_4{n}}{\sqrt{i}}$ which is equal to $\sqrt{1} + \sqrt{2} + \dots + \sqrt{\log_4{n}}$.

Upper bound
$\sqrt{1} + \sqrt{2} + \dots +\sqrt{\log{n}} \leq \log{n}\sqrt{\log{n}}$ which is less than $\sqrt{n}$ for sufficiently large $n$s (you may prove it using induction or using some other techniques).