Solve the recurrence relation:$ T(n) = \sqrt{n} T \left(\sqrt n \right) + n$ $$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$
Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.
The answer is $Θ (n \log \log n)$.
Can anyone arrive at the solution.

Use substitution method:
$$\Large\begin{align*} \text{T}(n) &= \sqrt{n}\ \text{T}(\sqrt{n})+n\\ &= n^{\frac{1}{2}}\ \text{T}\left(n^{\frac{1}{2}} \right )+n\\ &= n^{\frac{1}{2}}\left( n^{\frac{1}{2^2}}\ \text{T}\left(n^{\frac{1}{2^2}} \right )+n^{\frac{1}{2}} \right )+n\\ &= n^{\frac{1}{2}+\frac{1}{2^2}}\ \text{T}\left(n^{\frac{1}{2^2}}\right ) +n^{\frac{1}{2}+\frac{1}{2}}+n\\ &= n^{\frac{1}{2}+\frac{1}{2^2}}\ \text{T}\left(n^{\frac{1}{2^2}}\right ) +2n\\ &= n^{\frac{1}{2}+\frac{1}{2^2}}\left(n^{\frac{1}{2^3}}\ \text{T}\left(n^{\frac{1}{2^3}}\right ) +n^{\frac{1}{2^2}} \right )+2n\\ &= n^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}}\ \text{T}\left(n^{\frac{1}{2^3}}\right ) +n^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^2}} +2n\\ &= n^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}}\ \text{T}\left(n^{\frac{1}{2^3}}\right ) +3n\\ \vdots \\ &= n^{\sum_{i=1}^{k}\frac{1}{2^i}}\ \text{T}\left(n^{\frac{1}{2^k}}\right ) +kn\\ \end{align*}$$
assuming $\text{T}(2) = 2$, which is the least value of n that could be. So, $$\begin{align*} n^{\frac{1}{2^k}} &= 2\\ \frac{1}{2^k}\log_2(n) &= \log_2(2) \\ \log_2(n) &= {2^k} \\ \log_2\log_2(n) &= k\log_2(2) \\ \log_2\log_2(n) &= k \end{align*}$$
therefore, the recurrence relation will look like:
$$\large \begin{align*} \text{T}(n)&=n^{\sum_{i=1}^{k}\frac{1}{2^i}}\ \text{T}\left(n^{\frac{1}{2^k}}\right ) +kn\\ &=n^{\sum_{i=1}^{\log_2\log_2(n)}\frac{1}{2^i}}\ \text{T}\left(n^{\frac{1}{2^{\log_2\log_2(n)}}}\right ) +n \log_2\log_2(n)\\ \end{align*}$$
where, $$\sum_{i=1}^{\log_2\log_2(n)}\frac{1}{2^i} = 1-\frac{1}{\log_2(n)} = \text{fraction always, as }n\geq 2$$
so, $$\large \text{T}(n) = \mathcal{O}\left( n \log_2 \log_2 n \right)$$
