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Let $(a_n)_{n\geq 1}$ be a sequence of positive real numbers such that $a_n\rightarrow 0$ and $\sum\limits_{i=1}^{\infty}a_i$ is divergent. Prove that the set containing the sums of all $finite$ subsequences of $a_n$ is dense in $[0,\infty)$.

I know that a similar result holds. More exactly, under the same hypothesis, it was proved that the set containing the sums of $all$ subsequences (so, besides the finite subsequences, we also have the infinite ones) of $a_n$ is actually the interval $[0,\infty)$. Any hint/suggestion is appreciated.

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Well, you have almost everything you need. Let $x\in[0,\infty)$ and let $\epsilon>0$. We want to show that there is a finite subsequence of $(a_n)$ such that the sum of its elements is in the interval $(x-\epsilon, x+\epsilon)$. As you already know $x$ is the sum of all elements in a subsequence $(a_{n_k})$ of $a_n$. If the subsequence is finite then we are done because $\sum a_{n_k}=x\in(x-\epsilon, x+\epsilon)$ and that's what we need. Now suppose the subsequence is infinite. We still know that $\sum_{k=1}^\infty a_{n_k}=x$. An equivalent way to write it is $\lim_{M\to\infty}\sum_{k=1}^M a_{n_k}=x$. By the definition of the limit we can pick a large enough $M$ such that $\sum_{k=1}^M a_{n_k}\in(x-\epsilon,x+\epsilon)$. And $(a_{n_k})_{k=1}^M$ is a finite subsequence of $(a_n)$.

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You mentioned a result that has been proved, but I don't think it has been; the set containing the sums of all subsequences cannot be equal to $[0,\infty)$, as there is no subsequence which sums to $0$. Hopefully just a typo on your part?

Other than that the proof idea is the exact same. Pick a positive real $r>0$, and show you can "greedily" pick a finite subsequence of $a_n$ that sums to something in the range $(r-\epsilon,r)$ for any $\epsilon>0$.

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  • $\begingroup$ You're right. I should have stated that the set containing the sum of the zero sequence and the sums of the subsequences of $a_n$ is $[0,\infty)$. $\endgroup$ – Muffin Dec 9 '18 at 20:32

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