Strictly increasing infinite sequence can not exist in the set $S_k:=\{\frac{1}{n_1}+\cdots+\frac{1}{n_k}\mid n_1,n_2,\cdots,n_k\in \mathbb{N}\}$ Consider the following problem:
For each positive integer $k$, let $S_k$ be the set of real numbers that can be expressed in the form
$$\frac{1}{n_1}+\frac{1}{n_2}+\ldots+\frac{1}{n_k}$$
where $n_1,n_2,\ldots,n_k$ are positive integers.
Prove that $S_k$ doesn't contain a strictly increasing infinite sequence.

The following is my attempt, but I am not sure if it is correct because this solution seems to be leaning towards real analysis, which I really did not expect while starting out on the problem(I expected it to be number theoretic in nature).
Let us fix a $k$ first. Let $(s_n)_{n\ge 1}$ be a increasing sequence in $S_k$.
With each $s_i$ in the sequence, we can associate a tuple $N_i:=(\frac{1}{n_{1,i}},\frac{1}{n_{2,i}},\cdots, \frac{1}{n_{k,i}})$ such that
$$\frac{1}{n_{1,i}}+\frac{1}{n_{2,i}}+\cdots+\frac{1}{n_{k,i}}=s_i$$
But $(N_i)_{i\ge 1}$ is a sequence in $I^k$ where $I=[0,1]$. Since $I^k$ is compact, $(N_i)$ must have a convergent sub-sequence. By re-indexing the sequence, we assume that $(N_i)$ is a converging sequence. But then $(\frac{1}{n_{i,j}})_{j\ge 1}$ must be converging for $1\le i\le k$. But any sequence in $\{\frac{1}{n}\mid n\in \mathbb{N}\}$ converges to either $0$ or $\frac{1}{n}$(by using a eventually constant sequence). The $(N_i)_{i\ge 1}$ must converge to some $k$-tuple $A:=(a_1,a_2,\cdots,a_k)$, where $a_i=0$ or $\frac{1}{m_i}$ for some $m\in \mathbb{N}$ for $1\le i\le k$. If $a_i=\frac{1}{m_i}$ for some $i$, then we must have $n_{i,j}=m_i$ for large enough $j$. Thus it is possible to pick a large enough $M$ such that the slots where $(n_{i,j})_{j\ge 1}$ is converging to some $\frac{1}{m_i}$, are all constant(hence equal to $\frac{1}{m_i}$) for $j\ge M$. Now consider $d_{\infty}(A,N_M)$ where $d_{\infty}$ denotes the $L^{\infty}$ metric on $\mathbb{R}^k$. Let $N_0$ be the largest of the $n_{i,M}$. Since $N_i\to A$, we must have $d_{\infty}(N_i,A)<\frac{1}{2N_0}$ for $i\ge M_1$. Let $M_2=\max(M_1,M)$. Then $s_{M_2}<s_M$, a contradiction because $s_i$ is a increasing sequence.
 A: I suggest to do with by induction.
The base case $k = 1$ is trivial.
The induction hypothesis is that there does not exist a strictly increasing infinite sequence in $S_k$.
Now assume that there exists a strictly increasing sequence $(x_m)$ in $S_{k+1}$. Each $x_m$ has the form $x_m = s_m + \dfrac{1}{n_m}$ with $s_m \in S_k$ and $n_m \in \mathbb N$. The sequence $\left(\dfrac{1}{n_m}\right)$ is a sequence in $(0,1]$, thus we may assume w.l.o.g. that $\dfrac{1}{n_m} \to t \in [0,1]$. If $\dfrac{1}{n_m}$ attains only finitely many values, then $t$ must be one of these values and we may w.l.o.g. assume that $\left(\dfrac{1}{n_m}\right)$ is constant. If $\dfrac{1}{n_m}$ attains infinitely many values, then we get $t = 0$ and we may assume w.l.o.g. that $\left(\dfrac{1}{n_m}\right)$ is strictly decreasing. In both cases $\left(\dfrac{1}{n_m}\right)$  is non-increasing. This implies that $(s_m)$ must be strictly increasing in order that $(x_m)$ is strictly increasing. This contradicts the induction hypothesis.
