The recurrence relation for this is following:
$T(n)=T(n^{1/2})+n$ ......(1)
put $n=n^{1/2}$
we get,
$T(n^{1/2})=T(n^{1/4})+n^{1/2}$
put this in (1), we get,
$T(n)=T(n^{1/4})+n^{1/2}+n$
similarly,
$T(n)=T(n^{1/2^{k}})+n^{1/2^{k-1}}+......+n$
This will stop when $n^{1/2^{k}}=2$
Taking log of base 2 both side,we get
$1/2^{k} logn=log2
logn=2^{k}
log log n=k$
T(n)=O(log log n)