Why is $p_n\sim\sum_{k=1}^{n}\log(p_k)$? Why is
$$
p_n\sim\sum_{k=1}^{n}\log(p_k)
$$
where $p_n$ is the $n$th prime?
In addition, is it true that
$$
n\log\left(\dfrac{\sum_{k=1}^{n}\log(k)}{\log(\log(n))}\right)
$$
is a better approximation for $p_n$ than $n\log(n)$?
 A: For a quick qualitative answer, we note that by the prime number theorem $p_k \sim k\log k$, and hence we have $\log p_k \sim \log k + \log \log k \sim \log k$, and
$$\sum_{k=1}^n \log k \sim n \log n.$$
One has to do some work, however, to see that the summation of asymptotically equal terms leads in this case to asymptotically equal sums.
Probably more instructive is a summation by parts,
$$\begin{align}
\sum_{k=1}^n \log p_k &= \sum_{m \leqslant p_n} \left(\pi(m) - \pi(m-1)\right)\log m\\
&= \pi(p_n)\log p_n - \sum_{m\leqslant p_n} \pi(m)\left(\log (m+1) - \log m\right)\\
&= n\log p_n - \sum_{m\leqslant p_n} \pi(m) \left(\frac{1}{m} + O\left(m^{-2}\right)\right)\\
&= n\log p_n - \sum_{m\leqslant p_n} \frac{\pi(m)}{m} + O(\log\log p_n)\\
&= n\log p_n + O\left(\frac{p_n}{\log p_n}\right),
\end{align}$$
and the prime number theorem says that $n\log p_n \sim p_n$.
Regarding the addition, Raymond Manzoni reports in his answer to a question on bounds for $p_n$ that Pierre Dusart proved that
$$\begin{align}
\frac{p_n}n &\leqslant \log n+\log (\log n) -1+\frac{\log (\log n)-2}{\log n}\quad\text{for}\ n\geqslant 688383,\\
\frac{p_n}n &\geqslant\log n+\log (\log n)-1+\frac{\log (\log n)-21/10}{\log n}\quad\text{for}\ n\geqslant 3.
\end{align}$$
Since
$$\sum_{k=1}^n \log k = \int_1^n \log t\,dt + \sum_{k=2}^n \left(\log k - \int_{k-1}^k \log t\,dt\right) = n\log n - n + O(\log n),$$
we have
$$\begin{align}
\log \left(\frac{\sum_{k=1}^n \log k}{\log \log n}\right)
&= \log \left(\frac{n\log n - n + O(\log n)}{\log \log n}\right)\\
&= \log \left(n\frac{\log n - 1 + O\left(\frac{\log n}{n}\right)}{\log \log n}\right)\\
&= \log n + \log \left(\log n - 1 + O\left(\tfrac{\log n}{n}\right)\right) - \log \log \log n\\
&= \log n + \log \log n + O\left(\log \log \log n\right),
\end{align}$$
which means it is indeed a better approximation than $n\log n$. However, a still better approximation, at least for large $n$, is
$$p_n \approx n \log \left(\sum_{k=1}^n \log k\right)$$
without dividing the argument of the logarithm by $\log \log n$. Or the even more accurate $p_n \approx n(\log n + \log \log n - 1 + o(1))$ which is simple enough to remember even when one hasn't Dusart's bounds handy to look up.
