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We know that the sum of the reciprocals of all prime numbers diverges (that is, $\displaystyle\sum_{p\text { is prime}}p^{-1}=\infty$) and the sum of the reciprocals of all perfect squares converges (that is, $\displaystyle\sum_{k=1}^{\infty}k^{-2}<\infty$). Does this result imply that the set of prime numbers is more "dense" than the set of perfect squares; that is, given a sufficiently large integer $N$, there are always more primes than perfect squares in the closed interval $[1,N]$? If so, how about the answer to the following generalised question:

For a fixed positive number $\varepsilon$, does there exist a positive integer $N_{\varepsilon}$ such that if $N>N_{\varepsilon}$, there are always more primes than numbers of the form $k^{1+\varepsilon}$ (where $k$ is a positive number) in the closed interval $[1,N]$?

My guess is that the answer appears to be yes, since $\displaystyle\sum_{p\text { is prime}}p^{-1}=\infty$ whilst $\displaystyle\sum_{k=1}^{\infty}k^{-1-\varepsilon}<\infty$.

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    $\begingroup$ Ever heard of the Prime number theorem? Yes, your guess is true. $\endgroup$ – Ivan Neretin Mar 1 '18 at 12:05
  • $\begingroup$ Thanks @IvanNeretin! I was not aware of it. $\endgroup$ – Zuriel Mar 1 '18 at 15:12
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Your guessed right!

According to the prime number theorem, the number primes less than $N$ grows approximately at the rate $N/\ln N$ where as the number of numbers of the form $k^{1+\epsilon}$ which are less than $N$ grow at the rate $N^{1/(1+\epsilon)}$. Thus clearly number of primes grow faster because beyond a certain point, for every $\epsilon$, $\ln N$ will eventually grow slower than $N^{\epsilon/(1+\epsilon)}$.

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  • $\begingroup$ Thank you for your answer! Can this result be proven using the fact that the sum of reciprocals of all primes diverges but the sum of $1/k^{1+\varepsilon}$ converges? $\endgroup$ – Zuriel Mar 2 '18 at 3:08
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You can also use Bertrand's postulate, which says there is always at least one prime between $n$ and $2n - 2$ for $n > 3$.

Between $1$ and $n^2$ there are obviously $n$ squares (without loss of generality, we stipulate that $n$ is positive). Thanks to Bertrand, we can deduce that there is at least one prime between each multiple of $n$ and the next, from $n$ up to $n^2$. And if $n > 3$, then there are at least two primes between $1$ and $n$, which tells us there are at least $n + 1$ primes between $1$ and $n^2$.

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  • $\begingroup$ Thanks! But Bertrand's postulate will not work for the generalised question (the one compare number of primes and number of numbers of the form $k^{-1-\varepsilon}$), right? $\endgroup$ – Zuriel Mar 3 '18 at 1:02

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