I'm trying to bound the number of operations for computing the product of the first $n$ primes:
$$\prod_{i=1}^n p_i,\ \text{where}\ p_i\ \text{is the}\ i\text{th prime}$$
Since the $n$th prime is close to $n\log{n}$, its size in bits will be $\log{(n\log{n})}=\log{n}+\log{\log{n}}=O(\log{n})$ bits.
Given recent advances, let's assume that $k$-bit integers can be multiplied in $O(k\log{k})$ time.
Then, the time to multiply the first $n$ primes should be (roughly):
$$\sum_{i=1}^n |p_i|\log{|p_i|} = \sum_{i=1}^n \log{i}\log{\log{i}}$$
The best I can do to bound this is $O(n\log^2{n})$ time as follows: \begin{align*} \sum_{i=1}^n \log{i}\log{\log{i}} &\le \sum_{i=1}^n \log{i}\log{i} = \sum_{i=1}^n \log^2{i} \le \sum_{i=1}^n \log^2{n} = O(n\log^2{n}) \end{align*}
Yet, in some academic literature I'm reading, folks seem to be assuming this can be done in $O(n\log{n})$ time.
So, my question is: Is there a better bound on the above sum? (Or, did I mess up the analysis of the time?)
The best upper-bound I can come up with is actually $O(n\log{n}\log{\log{n}})$:
\begin{align*} \sum_{i=1}^n \log{i}\log{\log{i}} \le \log{\log{n}}\sum_{i=1}^n \log{i} \le \log{\log{n}}\sum_{i=1}^n \log{n} = O(n\log{n}\log{\log{n}}) \end{align*}
Then, @SandeepSilwal showed a lower-bound of $\Omega(n\log{n}\log{\log{n}})$ as follows: \begin{align*} \sum_{i=1}^n \log{i} \log{\log{i}} &\ge \sum_{i=n/2}^n \log{i} \log{\log{i}}\ge \log{\log{n/2}}\sum_{i=n/2}^n \log{i} \\ &\ge n/2 \log{n}\log{\log{n/2}}=\Omega(n\log{n}\log{\log{n}}) \end{align*}
Thus, it follows that multiplying the first $n$ primes takes $\Theta(n\log{n}\log{\log{n}})$ time.