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Let $(a_n)_{n\geq 1}$ be a sequence of non-negative real numbers such that $a_n \to 0$ but $\sum_{n\geq 1} a_n$ diverges. Show that the set of sums $\sum_{n \in S} a_n$, where $S$ ranges over the finite sets of positive integers, is dense, i.e. every open interval of non-negative real numbers contains a number which equals at least one of these sums.

I guess given a real $x$ one can greedily pick terms of the sequence to reach (as a sum) $x$ arbitrarily close, but no idea how to formalize this. Any help appreciated!

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    $\begingroup$ You might find the following paper interesting: Rafe Jones, Achievement sets of sequences, Amer. Math. Monthly 118 (2011), no. 6, 508-521. $\endgroup$
    – Dan Velleman
    Commented Aug 11, 2021 at 20:25

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Let the interval be $(x,y)$.

Let $b_i$ be the subsequence of the $a_i$ with $ a_i < y-x$.

$b_i$ excludes at most a finite number of $a_i$ and therefore its sums also diverge.

Now let $n$ be the largest value with $\sum\limits_{i=1}^n b_i \leq x$ and note that $x<\sum\limits_{i=1}^{n+1}b_i < x + (y-x) = y$.

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  • $\begingroup$ Why have you written $0<a_i<y-x$ ? You could write $0\leqslant a_i<y-x$. Why do you need that $0<a_i$ ? Moreover, why have you written that $b_i$ excludes at most a finite number of non-zero $a_i$ ? You could write that $b_i$ excludes at most a finite number of $a_i$. Why do you need to write “non-zero” ? $\endgroup$
    – Antonio
    Commented Aug 11, 2021 at 21:37

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