The Cardinality of the Set $\{\lfloor\frac{N}{i^k}\rfloor \ | \ i \in \mathbb{Z}^+\}$ Given $k$

Given a fixed real number $$k$$, I'd like to ask how to calculate the asymptotic behavior of the cardinality of the set $$S_k=\left\{\left\lfloor\frac{N}{i^k}\right\rfloor \ \bigg| \ i \in \mathbb{Z}^+\right\}$$ for sufficiently large $$N$$, here $$\lfloor\cdot\rfloor$$ is the floor function.

An example: when $$k=1$$, it turns out that $$|S_k|=\Theta(N^{\frac{1}{2}})$$.

It seems that $$|S_k|=\Theta(N^{\frac{1}{k+1}})$$ asymptotically, but I cannot prove it rigorously.

Write $$S_k=S_k^-\cup S_k^+$$, where $$S_k^+=\{\lfloor N/i^k\rfloor :i\in [1,N^{1/(k+1)})\cap \mathbb Z\}\\ S_k^-=\{\lfloor N/i^k\rfloor :i\in (N^{1/(k+1)},N^{1/k}]\cap \mathbb Z\}$$
• Obviously, $$|S_k^+|\in O(N^{1/(k+1)})$$, since there $$\lfloor N^{1/(k+1)}\rfloor$$ possibilities for $$i$$. With some more thought, the values $$\lfloor N/i^k\rfloor$$ are distinct for distinct $$i$$, so $$|S_k^+|\in \Theta(N^{1/(k+1)})$$.
• Furthermore, $$|S_k^-|\in O(N^{1/(k+1)})$$, since $$S_k^-\subseteq [1,N^{1/(k+1)})\cap \mathbb Z$$. You can in fact show the other inclusion by proving that for $$i\ge N^{1/(k+1)}$$, that $$\frac{N}{(i+1)^k}\ge \frac{N}{i^k}-1$$ so that as $$i$$ increases from $$N^{1/(k+1)}$$ to $$N^{1/k}$$, it hits every element of $$[1,N^{1/(k+1)})$$ in decreasing order.