The average of all primes is $$\lim\limits_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \text{prime}(k) ,$$ which diverges.
What is the smallest $r$ such that for $t>r$, $$\lim_{n \to \infty} \frac{1}{n^t} \sum_{k=1}^n \text{prime}(k)$$ converges.
And what is $\lim\limits_{n \to \infty} \frac{1}{n^4} \sum\limits_{k=1}^{n} \text{prime}(k)$?
First, note that
$$n\log n=\omega(n); \tag{1}$$ $$n\log n= o(n^{1+\epsilon})\; \forall\epsilon>0 \tag{2}$$
Furthermore, by Riemann summation,
$$\sum_{k=1}^n\;k^s =n^{s+1}\sum_{k=1}^n\frac{1}{n}(k/n)^s\sim n^{1-s}\int_0^1x^sdx=\frac{n^{s+1}}{s+1}\;\forall s>-1 \tag{3}$$
Next, with $p_n\sim n\log n $ and the above we may say
$$\frac{A}{2}n^2\sim A\sum_{k=1}^n k \le \sum_{k=1}^n\;p_k\le B\sum_{k=1}^n k^{1+\epsilon}\sim\frac{B}{\epsilon+2} n^{\epsilon+2}. \tag{4}$$
(For some $A,B>0$, any $\epsilon>0$, and sufficiently large $n$.) Hence $$\frac{1}{n^t}\sum_{k=1}^n\; p_k$$ is asymptotically greater than a constant times $n^{2-t}$ (so $\lim$ diverges for $t<2$), and is also asymptotically less than a constant times $n^{2-t+\epsilon}$ for any $\epsilon>0$ (so $\lim=0$ for $t>2$).
When $t=2$, use Abel's summation formula to say (for some $C>0$) $$\sum_{k=1}^n\; p_k\ge C\sum_{k=1}^n k\log k= \frac{n(n+1)}{2}\log n-\int_1^n \frac{\lfloor x\rfloor(\lfloor x\rfloor+1)}{2}\frac{1}{x}dx$$ $$=\frac{n(n+1)}{2}\left(\log n - O(1)\right) =\omega(n^2)$$ hence $\lim$ doesn't exist at $t=2$.
Lets try to be even more precise, and work out the asymptotic. In this answer: How does $ \sum_{p<x} p^{-s} $ grow asymptotically for $ \text{Re}(s) < 1 $? It is shown that for fixed $s$ $$\sum_{p\leq x} p^{-s} \sim \frac{x^{1-s}}{(1-s)\log x}.$$ Letting $s=-1$ we see that $$\sum_{p\leq x} p \sim \frac{x^{2}}{2\log x}.$$ Now, the sum you are looking at is $$\sum_{k=1}^n p_k = \sum_{p\leq p_n} p \sim \frac{(p_n)^2}{2 \log (p_n)}.$$ Using the fact that $$p_n \sim n\log n$$ we see that
$$\sum_{k=1}^n p_k \sim \frac{n^2 \log n}{2}.$$
From here we can easily conclude all the other observations.
I hope that helps,
