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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 $\lim\limits_{n \to \infty} \frac{1}{n^4} \sum\limits_{k=1}^{n} \text{prime}(k)$?

3 Answers 3

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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 $t=2$.

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Your limit doesn't exist for $t=2$, and is zero for all $t\gt2$.