Why is $p_n\sim\sum_{k=1}^{n}\log(p_k)$?

Question

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 $n\log(n)$?

$p_n \sim n\log n$, so $\log p_k \sim \log k$, and $\sum_{k=1}^n \log k \sim n\log n$. — Daniel Fischer, Aug 02 '14 at 17:46
Seems to me that if it is true for all $P_n$ primes, then it has to be true at least for their "neighbors". — barak manos, Aug 02 '14 at 17:46
@DanielFischer Wouldn't $\log p_k \sim k\log k$ instead of just $\log k$? — Darth Geek, Aug 02 '14 at 17:49
@DarthGeek $\log p_k \approx \log (k\log k) = \log k + \log \log k$, and the $\log\log k$ term is asymptotically negligible. — Daniel Fischer, Aug 02 '14 at 17:51
Is it licit to sum equivalents ? In my memory it is not obviously true... — D.L., Aug 02 '14 at 17:56
Why if $u_n\simeq v_n$ and $u'_n\simeq w_n$, $u_n+u'_n\simeq v_n+w_n$ ? (I don't find the good command sorry...) — D.L., Aug 02 '14 at 18:04
More accurate, why do we have $\sum_{k=1}^n p_k \simeq \sum_{k=1}^n log(k)$ ? — D.L., Aug 02 '14 at 18:14
Sorry, I have understood my problem. The point is that the sum of equivalents is licit in this case because it is the partial sum of a nonconvergent infinite sum, so there's no problem. — D.L., Aug 02 '14 at 18:16

score 8 · Accepted Answer · edited Apr 13 '17 at 12:20

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.

thank you for the detailed answer - this is much clearer - again, many thanks — martin, Aug 02 '14 at 18:26
I only just saw the addition, I've added an answer to that too. — Daniel Fischer, Aug 02 '14 at 20:59

Why is $p_n\sim\sum_{k=1}^{n}\log(p_k)$?

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