# A question on the sums of finite subsequences of a sequence of positive reals

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.

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 $$(a_n)$$.
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 $$\epsilon>0$$.
• 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 $[0,\infty)$. – Muffin Dec 9 '18 at 20:32