Is the average of primes up to $n$ smaller than $\frac{n}{2}$, if $n > 19$? For $n = 19$, the average $\frac{(2 + 3 + 5 + 7 + 11 + 13 + 17 + 19)}{ 8} = \frac{77}{8}$ is greater than $\frac{19}{2}$ but for $n > 19$, it seems that the average is smaller than $n/2$.
I confirmed it holds for all $n < 10^7$ but have not been able to prove it for all n. How can I prove this?
 A: The answer is yes for sufficiently large $n$. Sketch of proof:

*

*Using summation by parts (as in either one of these two existing answers), the average in question is
$$
\frac1{\pi(n)}\sum_{p\le n} p = \frac1{\pi(n)} \biggl( n\pi(n) - \sum_{k<n} \pi(k) \biggr) = n - \frac1{\pi(n)} \sum_{k<n} \pi(k).
$$

*Instead of just using $\pi(x) = \dfrac x{\log x} + O\biggl( \dfrac x{\log^2 x} \biggr)$, we use the more precise
$$
\pi(x) = \dfrac x{\log x} + \dfrac x{\log^2 x} +O\biggl( \dfrac x{\log^3 x} \biggr)
$$
that follows from the asymptotic expansion of the logarithmic integral.

*If we write
$\displaystyle
\pi(n) = \dfrac n{\log n} \biggl( 1 + \dfrac 1{\log n} + O\biggl( \dfrac1{\log^2 n} \biggr) \biggr)
$
and use $(1+\varepsilon)^{-1} = 1-\varepsilon+O(\varepsilon^2)$, we see that
$$
\frac1{\pi(n)} = \dfrac{\log n}n \biggl( 1 - \dfrac 1{\log n} + O\biggl( \dfrac1{\log^2 n} \biggr) \biggr).
$$

*Since the function $\dfrac x{\log^\alpha x}$ is increasing for $x>e^\alpha$, it's easy to see that $\displaystyle \sum_{k<n} \frac k{\log^\alpha k} = \int_2^n \frac t{\log^\alpha t}\,dt + O(n)$.

*Integration by parts yields $\displaystyle \int_2^n \frac t{\log^\alpha t}\,dt = \frac{n^2}{2\log^\alpha n} + O(1) + \frac{\alpha}2 \int_2^n \frac t{\log^{\alpha+1} t}\,dt$. It follows that
$\displaystyle
\int_2^n \frac t{\log^2 t}\,dt = \frac{n^2}{2\log^2 n} + O\biggl( \frac{n^2}{\log^3 n} \biggr)
$ and
$\displaystyle\int_2^n \frac t{\log t}\,dt = \frac{n^2}{2\log n} + \frac{n^2}{4\log^2 n} + O\biggl( \frac{n^2}{\log^3 n} \biggr)
$.

Putting these all together:
\begin{align*}
\frac1{\pi(n)} \sum_{p\le n} p &= n - \frac1{\pi(n)} \sum_{k<n} \biggl( \dfrac k{\log k} + \dfrac k{\log^2 k} + O\biggl( \dfrac k{\log^3 k} \biggr) \biggr) \\
&= n - \frac1{\pi(n)} \biggl( \int_2^n \frac t{\log t}\,dt + \int_2^n \frac t{\log^2 t}\,dt + O\biggl( \int_2^n \frac t{\log^3 t}\,dt + n\biggr) \biggr) \\
&= n - \dfrac{\log n}n \biggl( 1 - \dfrac 1{\log n} + O\biggl( \dfrac1{\log^2 n} \biggr) \biggr) \\
&\qquad{}\times\biggl( \biggl( \frac{n^2}{2\log n} + \frac{n^2}{4\log^2 n} \biggr) + \frac{n^2}{2\log^2 n} + O\biggl( \frac{n^2}{\log^3 n} \biggr) \biggr) \\
&= n - \biggl( \frac n2 + \frac n{4\log n} + O\biggl( \frac n{\log^2n} \biggr) \biggr) = \frac n2 - \frac n{4\log n} + O\biggl( \frac n{\log^2n} \biggr),
\end{align*}
which finishes the proof.
A: Mandl's inequality (see for example this paper of Axler) states that
$$
\frac{1}{m}\sum\limits_{k = 1}^m {p_k }  < \frac{{p_m }}{2}
$$
for $m\geq 9$. If $n\geq 23$, then $\pi(n)\geq 9$. Thus,
$$
\frac{1}{{\pi (n)}}\sum\limits_{p \le n} p  = \frac{1}{{\pi (n)}}\sum\limits_{k = 1}^{\pi (n)} {p_k }  < \frac{{p_{\pi (n)} }}{2} \le \frac{n}{2}.
$$
For $n=20,21$ and $22$, the inequality can be checked by hand.
