= Solution
Use <natural logarithms>; the same argument works for any fixed base greater than $1$. The <function> $\log\log x$ is increasing for $x>1$, since its <derivative> is $1/(x\log x)>0$. With natural logs the sole negative summand is at $i=2$. Put $C_0=|\log\log2|$.
For all sufficiently large $n$ with $\log\log n\geq1$, monotonicity and the <triangle inequality> give
$$
\left|\sum_{i=2}^n\log\log i\right|
\leq C_0+\sum_{i=3}^n\log\log i
\leq C_0+n\log\log n
\leq(1+C_0)n\log\log n.
$$
These are fixed constants and an eventual bound, proving
$$
\boxed{\sum_{i=2}^n\log\log i=O(n\log\log n)}.
$$
For another fixed logarithm base there are still only finitely many negative initial terms, which are absorbed in $C_0$.
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