It is well known that, where $p_k$ is the $k$th prime number (this is $2 = p_1 < p_2 < p_3 < \cdots$), the following
Proposition. The series of reciprocals of primes $$\sum_{k=1}^\infty \frac{1}{p_k}$$ diverges.
Too is known the so called harmonic-arithmetic mean inequality
Proposition. For any positive real numbers $a_1,a_2,\ldots,a_N$, we have $$\left(a_1+a_2+\cdots +a_N\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_N}\right)\geq N^2.$$
Putting $a_k=p_k$ for $1\leq k\leq N$ we obtain $$0<\frac{N^2}{\sum_{k=1}^N p_k}\leq \sum_{k=1}^N \frac{1}{p_k}.$$
Thus $$\lim_{N\to \infty}\frac{N^2}{\sum_{k=1}^N p_k}\leq \infty.$$
My
Question. a) Compute previous limit. b) Compute $\lim_{N\to \infty} \frac{N^\alpha\cdot (\log N)^\beta}{\sum_{k=1}^N p_k}$, where I repeat another time that $p_k$ is the kth prime number and $\alpha,\beta$ real parameters.
Thanks in advance, my only goal is learn in this site Math Stack Exchange from yours answers.
References:
If you want read it, you could find references of proofs of divergence of reciprocals of primes and a proof of the harmonic-arithmetic mean inequality, via this web site or Wikipedia, for example.
The sum of the first $N$ primes is asymptotically equal to $\frac12 N^2 \log N$.
Your limit in a) is thus $0$.
For b) it is $0$ for $\alpha < 2$ and any $\beta$ and $\infty$ for $\alpha >2$ and any $\beta$.
If $\alpha= 2$ then it is $0$ for $\beta < 1$, $\infty$ for $\beta >1$ and $2$ for $\beta =1$.