TITLE: For what $t$ does $\lim\limits_{n \to \infty} \frac{1}{n^t} \sum\limits_{k=1}^n \text{prime}(k)$ converge?
QUESTION [10 upvotes]: 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 $\lim\limits_{n \to \infty} \frac{1}{n^4} \sum\limits_{k=1}^{n} \text{prime}(k)$?

REPLY [4 votes]: 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,