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Consider the set $\{\frac{1}{n} + \frac{1}{m}: n,m \in \mathbb{N} \}$. Is this set dense in some interval of $\mathbb{R}$?

More generally let $S_k= \{\sum_{i=1}^k \frac{1}{n_i}: n_i \in \mathbb{N} \} $. Is this set dense in some interval of $\mathbb{R}$?

I think the answer to both questions is no, but I was not able to come up with a proof. I am a physics student so I apologise if this question is naive for this website.

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We may assume $n\le m$ throughout.

Let $y>x>0$. There are only finitely many $n$ with $$ 1/n\ge x/2. $$ For each such $n$, the set of $1/n+1/m$ as $m$ varies is not dense in any subinterval of $(x,y)$.

A finite union of sets not dense in any subinterval of $(x,y)$ is still not dense in any subinterval of $(x,y)$, since you can "escape" the sets one at a time.

Hence your set is not dense in $(x,y)$.

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    $\begingroup$ The same argument shows that for any sequences $(u_n),(v_n)$ tending to zero (and with infinite image), the only limit points of $\{u_n+v_m,n,m\ge 0\}$ are $\{0\}\cup\{u_n:n\ge 0\}\cup\{v_n:n\ge 0\}$ (and same for finitely many sequences). $\endgroup$
    – YCor
    Commented Nov 15, 2019 at 9:04
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The answer is no. Because any convergent sequence of elements of $S_k$ has rational limit. Let $s_p$ be such a sequence. Then we have sequence $q_p=(n_{1p};\dots;n_{kp})\in\mathbb{N}^k$ such that $s_p=\sum_i \frac{1}{n_{ip}}$. Let $T=\{i_1;\dots;i_r\}$ is the set of indices such that the sequence $\{n_{i_kp}\}_{p\in\mathbb{N}}$ is stationary beginning from some $p_{i_k}$. Then $$ \lim_{p\to\infty}s_p = \begin{cases} \sum\limits_{i\in T}\frac{1}{n_{ip_{i}}},~\text{if}~~ T\neq\emptyset \\ 0,~\text{if}~~ T=\emptyset \end{cases} $$

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