How to get asymptotic behaviour for this function?

Question

Let $$F(N)=\prod_{p\ \text{prime}}\left(1-\frac{1}{N^2}{\left\lfloor{\frac{N}{p}}\right\rfloor}^2\right)$$ My claim is that, in the limit as $N$ goes to $+\infty$, this function approaches $\frac{6}{\pi^2}$. This is motivated by the fact that $$\frac{1}{\zeta(s)}=\prod_{p\ \text{prime}}\left(1-\frac{1}{p^s}\right)$$ $$\implies\ \frac{1}{\zeta(2)}=\frac{6}{\pi^2}=\prod_{p\ \text{prime}}\left(1-\frac{1}{p^2}\right)$$ However, I think there is a problem with the floor function. In my claim I am assuming that, as $N$ gets large, I can "simplify" the $\frac{1}{N^2}{\left\lfloor{\frac{N}{p}}\right\rfloor}^2$ to just $\frac{1}{p^2}$, since as $N$ grows ${\left\lfloor{\frac{N}{p}}\right\rfloor}\sim \frac{N}{p}$. Is this correct? If not, can we get a value for the limit, or an asymptotic behaviour for $F(N)$?

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This question is motivated by the fact that, if the claim is true, we have an elementary proof of the result $$\sum_{k=1}^n \varphi(k)\simeq \frac{3}{\pi^2}n^2$$ which, I believe, is usually proved only with complicated number theory methods. For those who would like to learn more, here is the link to the original question.

Answer 1

You say this result would lead to an elementary proof that $\sum_{k\leq n} \varphi(k) \sim (3/\pi^2)n^2$, but that argument would still require knowing $\zeta(2) = \pi^2/6$, so the proof would not really be elementary.

For $x \geq 1$, set $\Phi(x) = \sum_{n \leq x} \varphi(n)$. Your premise is that showing $\Phi(x) \sim (3/\pi^2)x^2$ needs complicated methods. I disagree. Below is an argument that $\Phi(x) = (3/\pi^2)x^2 + O(x\log x)$. (Perhaps if you demand sharper error terms the methods will start looking more complicated.)

We start by writing $\varphi(k)$ as a sum: $\varphi(k) = k\prod_{p \mid k} (1 - 1/p) = \sum_{d \mid k} k\mu(d)/d$. (An alternative method: $\sum_{d \mid k} \varphi(d) = k$ for all $k \geq 1$, so $\sum_{d \mid k} \mu(d)k/d = \varphi(k)$ for all $k \geq 1$ by Moebius inversion.)

Thus \begin{eqnarray} \Phi(x) & = & \sum_{k \leq x} \sum_{d \mid k} \frac{k}{d}\mu(d) \nonumber \\ & = & \sum_{dd' \leq x} \mu(d)d' \nonumber \\ & = & \sum_{d \leq x} \mu(d)\sum_{d' \leq x/d} d' \nonumber \\ & =& \sum_{d \leq x} \mu(d) \frac{1}{2}\left(\left\lfloor\frac{x}{d}\right\rfloor^2 + \left\lfloor\frac{x}{d}\right\rfloor\right). \nonumber \end{eqnarray}

Since $\lfloor x/d\rfloor = x/d + O(1)$, $\lfloor x/d\rfloor^2 + \lfloor x/d\rfloor = x^2/d^2 + O(x/d)$ for $d \leq x$, so $$ \Phi(x) = \sum_{d \leq x} \mu(d)\left(\frac{x^2}{2d^2} + O(x/d)\right) = \frac{x^2}{2}\sum_{d \leq x} \frac{\mu(d)}{d^2} +O(x\log x). $$ The series $\sum_{d \geq 1} \mu(d)/d^2$ has value $1/\zeta(2) = 6/\pi^2$, so $$ \Phi(x) = \frac{x^2}{2}\left(\frac{6}{\pi^2} - \sum_{d > x} \frac{\mu(d)}{d^2}\right) + O(x\log x). $$ Now let's bound $|\sum_{d > x} \mu(d)/d^2|$. It is less than $\sum_{d > x} 1/d^2 < \int_{\lfloor x\rfloor}^\infty dt/t^2 = O(1/x)$ for $x \geq 1$, so
$$ \Phi(x) = \frac{3}{\pi^2}x^2 + x^2O(1/x)+ O(x\log x) = \frac{3}{\pi^2}x^2 + O(x\log x). $$