solving Recurrence relation f T(n)=T(n/2)+log log(n) T(n)=T(n/2)+log log(n)--------1
T(n)=T(n/4)+log log(n/2)+log log(n)-------------2
T(n)=T(n/8)+log log(n/4)+log log(n/2)+log log(n)-----------3
T(n)=T(n/2k)+log log(n/k)+....+log log(n/2)+log log(n)
T(n)=1+log log(n/k)+....+log log(n/2)+log log(n)
 n/2k=1 =&gt; n=2k
 T(n)=1+log log(2K/k)+....+log log(n/2)+log log(n)
 T(n)=1+log log(2)+....+log log(n/2)+log log(n)
 T(n)=1+log (log(2)*....*log(n/2)*log(n))

I ARRIVED AT THIS POINT BUT I DIDN'T KNOW HOW TO CONTENU ANY HELP PLEASE
Let's consider $n=2^k$:
$$\begin{array}{l}T(n)=T\left(\frac{n}{2}\right)+\log\log n = T\left(\frac{n}{2^2}\right)+\log\log \frac{n}{2}+\log\log n = \\
=T\left(\frac{n}{2^3}\right)+\log\log \frac{n}{2^2}+\log\log \frac{n}{2}+\log\log n=\\
=\cdots=\\
=T\left(\frac{n}{2^k}\right)+\log\log\frac{n}{2^{k-1}}+ \log\log\frac{n}{2^{k-2}}+\cdots+\log\log n=\quad(3)\\
=T(1)+\log\log 2 + \log\log 2^2 + \cdots + \log\log 2^k=\\
=T(1)+\log\left[ \log2\cdot \log2^2 \cdots  \log2^k \right] =\\
= T(1)+\log\left[ k!\log^k2 \right]=\\
=T(1) + \log k! +k \log\log 2 \in \Theta(k \log k)\end{array}$$
where $k=\log_2 n$.