Finding $\sum_{i=0}^{\log_2\log_2n} n^{\frac{2^{i+1}-1}{2^i}}$

Question

I have a task on finding runtime complexity of an algorithm. I worked out the following summation expression, unsuccessfully tried to get a result from Wolfram Alpha, have no idea in what direction I should search.

$$\sum_{i=0}^{\log_2\log_2n} n^{\frac{2^{i+1}-1}{2^i}}$$

Answer 1

Rewriting as

$$n^2\sum_{i=0}^{\log_2 \log_2 n} n^{-2^{-i}}$$

we can find lower and upper bounds by recognizing that

$$n^{-1} < n^{-2^{-i}} < n^{\frac{-1}{\log_2 n}} = n^{-\log_n 2} = 2^{-1}$$

which implies

$$ n \log_2 \log_2 n < n^2\sum_{i=1}^{\log_2 \log_2 n} n^{-2^{-i}} < \frac{1}{2}n^2 \log_2 \log_2 n$$