Cardinality of set of fractional sums What is the cardinality of the set $S_2$:
$$ \frac{1}{a_1^n} + \frac{1}{a_2^n}, 1 \leq a_1,a_2 \leq k \in N$$
for different values of $n$?
I suspect there is an $n_0$ for which $|S_2| = \binom{k+1}{2}, \forall n \geq n_0$. 
Is there such an $n_0$ for all sets $S_m$:
$$ \frac{1}{a_1^n} + \frac{1}{a_2^n} + \cdots + \frac{1}{a_m^n}, 1 \leq a_i \leq k$$
i.e. $n_0: |S_m| = \binom{k+m-1}{m}, \forall n \geq n_0$?
Example:
For the set $S_2$, with $k=3, n=2$:
$$ S = \{\frac{1}{1} + \frac{1}{1}, \frac{1}{1}+ \frac{1}{2^2}, \ldots, \frac{1}{3^2} + \frac{1}{3^2} \} = \{ 2, \frac{5}{4}, \frac{10}{9}, \frac{1}{2}, \frac{13}{36}, \frac{2}{9} \}$$
so $$ |S_2(k=3,n=2)| = 6 $$
 A: Proposition. For each $m, k\ge 2$, $|S_m(k,n)|={k+m-1\choose m}$ for each $n\ge \log_{\frac k{k-1}} m-1$.
Proof. We shall follow A.P.’s comment. Let $\mathcal S$ be a family of non-decreasing sequences $(a_1,\dots, a_m)$ of natural numbers between $1$ and $k$. For each $S=(a_1,\dots, a_m)\in \mathcal S$ put  $\Sigma S=a_1^{-n}+\dots a_m^{-n}$. Since $|\mathcal S|={k+m-1\choose m}$,  we have to show that numbers  $\Sigma S$ are distinct when $S\in\mathcal S$. Suppose to the contrary, that there exists distinct sequences $S=(a_1,\dots, a_m)$ and $T=(b_1,\dots, b_m)$ in $\mathcal S$ such that $\Sigma S=\Sigma T$. Let $i$ be the smallest number such that $a_i\ne b_i$. Without loss of generality we may suppose that $a_i<b_i$. Then 
$$\Sigma S-\Sigma T\ge a^{-n}_i- b^{-n}_i+\sum_{j>i} a^{-n}_j - b^{-n}_j\ge$$ $$a^{-n}_i- b^{-n}_i+\sum_{j>i} k^{-n} - b^{-n}_i=$$ $$a^{-n}_i+(m-i)k^{-n}-(m-i+1)b^{-n}_i\ge$$ $$ a^{-n}_i+(m-i)k^{-n}-(m-i+1)(a_i+1)^{-n}\ge$$ $$ a^{-n}_i+(m-1)k^{-n}-m(a_i+1)^{-n}=f(a_i),$$
where $f(x)=x^{-n}+(m-1)k^{-n}-m(x+1)^{-n}$. Since $f(k-1)=(k-1)^{-n}-k^{-n}>0$, to obtain a contradiction it suffices to show that 
$f’(x)<0$ for $1\le x<k-1$. Since $f’(x)=(-n)x^{-n-1}+nm(x+1)^{-n-1}$, we have to check that 
$x^{-n-1}-m(x+1)^{-n-1}>0$
$x^{-n-1}>m(x+1)^{-n-1}$
$\left( x+\frac{1}{x}\right)^{n+1}>m$
which is true, because $\left(x+\frac{1}{x}\right)^{n+1}>\left(1+\frac{1}{k-1}\right)^{n+1}\ge m$. $\square$
