Suppose \(\{a_n\}\) is a sequence of
positive real numbers such that \(\lim_{n\to\infty}a_n=0\) and \(\sum a_n\) diverges. Prove that for all
\(x>0\) there exist integers \(n(1)<n(2)<\ldots\) such that \(\sum_{k=1}^\infty a_{n(k)}=x\).
(Note: Many variations on this problem are possible including more general rearrangements. You may also wish to show that if \(\sum a_n\) converges conditionally then given any \(x\in\mathbb{R}\) there is a rearrangement of \(\{b_n\}\) of \(\{a_n\}\) such that \(\sum b_n=r\). See Rudin Thm. 3.54 for a further generalization.)
Solution
1. Construction: fix \(x > 0\); define \(n(1)\) as the smallest index with \(a_{n(1)} \le x\), and inductively \(n(k)\) as the smallest index \(> n(k-1)\) with \(a_{n(k)} \le x - \sum_{j<k}a_{n(j)}\) (stop if the sum equals \(x\)).
such indices exist at every step: if no index \(\ge n(k-1) + 1\) had \(a_n \le x - s_{k-1} > 0\), then all remaining terms would be \(> x - s_{k-1} > 0\), contradicting \(a_n \to 0\).
2. The partial sums \(s_k = \sum_{j \le k}a_{n(j)}\) are nondecreasing and bounded above by \(x\).
each added term satisfies \(a_{n(k)} \le x - s_{k-1}\), so \(s_k \le x\); hence \(s_k \uparrow s^* \le x\).
3. \(s^* = x\).
suppose \(s^* < x\); let \(\eta = x - s^* > 0\).
Since \(a_n \to 0\), eventually \(a_n < \eta\); and since \(s_k \to s^*\), eventually \(x - s_{k-1} > x - s^* = \eta\)… more precisely \(x - s_{k-1} \ge \eta/2 > \eta/2\) for \(k\) large. Then every term \(a_n\) beyond that point satisfies \(a_n < \eta/2 \le x - s_{k-1}\), so the greedy construction adds all remaining terms: \(s_k = s_N + \sum_{n=N}^{n(k)}a_n \to \infty\) (as \(\sum a_n\) diverges), contradicting \(s_k \le x\). Hence \(s^* = x\).
4. Q.E.D.
2 and 3 give \(\sum_{k=1}^\infty a_{n(k)} = x\).