 But if the terms \(a_n\) converge to 0, and each one is smaller in absolute value and has the opposite sign as the previous term, then the infinite sum must converge. If the first term in the sequence is positive, this is because the even numbered partial sums form a bounded, increasing sequence, and the odd numbered partial sums form a bounded, decreasing sequence, and the difference between successive terms of these two sequences is a term of the original sequence and thus converges to 0. Thus, the sum \(\sum_{i=1}^\infty\frac{(-1)^{i+1}}{i}\) converges (in particular, to the natural logarithm of 2). But a somewhat surprising fact, known as the Riemann rearrangement theorem, states that if the positive terms of a sequence have no finite sum, and the negative terms of the sequence have no finite sum, but the individual terms themselves converge to 0, then for any real number \(x\), the terms of the sequence can be put into some order so that \(x\) is the sum of the sequence in that order! To prove this, just rearrange the terms by beginning with enough positive terms to bring a partial sum above \(x\), then enough negative terms to bring a partial sum below \(x\), then enough partial sums to bring a partial sum above \(x\) again, and so on. This process must be able to be carried out, because the partial sum of the positive terms eventually exceeds any finite bound, and similarly with the partial sum of the negative terms. Since the individual terms of the sequence are all eventually within \(\epsilon\) of 0, these partial sums must eventually never differ from \(x\) by more than \(\epsilon\), and so the sum of this arrangement of the terms must converge to \(x\).