Find an upper bound for $T(n)=T(\sqrt{n})+10\log\log n$ I need to find an upper bound for $T(n)=T(\sqrt{n})+10\log\log n$.

I thought first to make a substitution: $m=\log n$. Then:
$$
T(2^m)=T(2^{m \over 2})+10\log m
$$
Let $S(m)=T(2^m)$:
$$
S(m)=S\big({m \over 2}\big)+10\log m
$$

Now we can use Master theorem. 

Suppose it's the case when $f(m)=O(m^{\log_2 1-\epsilon})=O(m^{0-\epsilon})$. Let $\epsilon=0.5$ then:
$$
\lim_{n\to \infty}\frac{10\log m}{m^{0.5}}=0
$$
Hence according to the Master theorem $T(m)=\Theta(m^0)=\Theta(1)$.

Am I in the right direction?
An alternative solution still using domain transformation/change of variables.

$$T(n) = T(\sqrt{n}) + \log \log n$$

1. Let $m = \log n$

We can then define a new function $S$ based on how $m$ changes in $T$ and expand it:

$$\begin{align}
S(m) &amp;= S\left(\frac{m}{2}\right) + \log m\\
&amp;=S\left(\frac{m}{4}\right) + \log m -1 + \log m\\
&amp; \vdots\\
&amp; = \log^2 m - \sum_{i = 1}^{\log m} i \\
&amp; = \Theta(\log^2 m) \implies T(n) = \Theta((\log \log n)^2) 
\end{align}$$

2. Let $k = \log \log n$

We can then define a new function $R$ based on how $k$ changes in $T$ and expand it:

$$\begin{align}
R(k) &amp;= R(k - 1) + k\\
&amp; = R(k - 2) + k - 1 + k\\
&amp; \vdots\\
&amp; = \Theta(k^2) \implies T(n) = \Theta((\log \log n)^2)
\end{align}$$