The simplest and most common conditionally convergent series are what is known as alternating series, series whose terms are alternately positive and negative. The convergence of the most important series of this type is established by the following theorem.

If $$\phi(n)$$ is a positive function of $$n$$ which tends steadily to zero as $$n \to \infty$$, then the series $\phi(0) – \phi(1) + \phi(2) – \dots$ is convergent, and its sum lies between $$\phi(0)$$ and $$\phi(0) – \phi(1)$$.

Let us write $$\phi_{0}$$, $$\phi_{1}$$, … for $$\phi(0)$$, $$\phi(1)$$, …; and let $s_{n} = \phi_{0} – \phi_{1} + \phi_{2} – \dots + (-1)^{n}\phi_{n}.$ Then $s_{2n+1} – s_{2n-1} = \phi_{2n} – \phi_{2n+1}\geq 0,\quad s_{2n} – s_{2n-2} = -(\phi_{2n-1} – \phi_{2n}) \leq 0.$ Hence $$s_{0}$$, $$s_{2}$$, $$s_{4}$$, …, $$s_{2n}$$, … is a decreasing sequence, and therefore tends to a limit or to $$-\infty$$, and $$s_{1}$$, $$s_{3}$$, $$s_{5}$$, …, $$s_{2n+1}$$, … is an increasing sequence, and therefore tends to a limit or to $$\infty$$. But $$\lim (s_{2n+1} – s_{2n}) = \lim (-1)^{2n+1} \phi_{2n+1} = 0$$, from which it follows that both sequences must tend to limits, and that the two limits must be the same. That is to say, the sequence $$s_{0}$$, $$s_{1}$$, …, $$s_{n}$$, … tends to a limit. Since $$s_{0} = \phi_{0}$$, $$s_{1} = \phi_{0} – \phi_{1}$$, it is clear that this limit lies between $$\phi_{0}$$ and $$\phi_{0} – \phi_{1}$$.

Example LXXVIII
1. The series $\begin{gathered} 1 – \frac{1}{2} + \frac{1}{3} – \frac{1}{4} + \dots,\quad 1 – \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} – \frac{1}{\sqrt{4}} + \dots,\\ \sum \frac{(-1)^{n}}{(n + a)},\quad \sum \frac{(-1)^{n}}{\sqrt{n + a}},\quad \sum \frac{(-1)^{n}}{(\sqrt{n} + \sqrt{a})},\quad \sum \frac{(-1)^{n}}{(\sqrt{n} + \sqrt{a})^{2}},\end{gathered}$ where $$a > 0$$, are conditionally convergent.

2. The series $$\sum(-1)^{n}(n + a)^{-s}$$, where $$a > 0$$, is absolutely convergent if $$s > 1$$, conditionally convergent if $$0 < s \leq 1$$, and oscillatory if $$s \leq 0$$.

3. The sum of the series of § 188 lies between $$s_{n}$$ and $$s_{n+1}$$ for all values of $$n$$; and the error committed by taking the sum of the first $$n$$ terms instead of the sum of the whole series is numerically not greater than the modulus of the $$(n + 1)$$th term.

4. Consider the series $\sum \frac{(-1)^{n}}{\sqrt{n} + (-1)^{n}},$ which we suppose to begin with the term for which $$n = 2$$, to avoid any difficulty as to the definitions of the first few terms. This series may be written in the form $\sum \left[\left\{ \frac{(-1)^{n}}{\sqrt{n} + (-1)^{n}} – \frac{(-1)^{n}}{\sqrt{n}}\right\} + \frac{(-1)^{n}}{\sqrt{n}}\right]$ or $\sum \left\{\frac{(-1)^{n}}{\sqrt{n}} – \frac{1}{n + (-1)^{n}\sqrt{n}}\right\} = \sum (\psi_{n} – \chi_{n}),$ say. The series $$\sum \psi_{n}$$ is convergent; but $$\sum \chi_{n}$$ is divergent, as all its terms are positive, and $$\lim n\chi_{n} = 1$$. Hence the original series is divergent, although it is of the form $$\phi_{2} – \phi_{3} + \phi_{4} – \dots$$, where $$\phi_{n} \to 0$$. This example shows that the condition that $$\phi_{n}$$ should tend steadily to zero is essential to the truth of the theorem. The reader will easily verify that $$\sqrt{2n + 1} – 1 < \sqrt{2n} + 1$$, so that this condition is not satisfied.

5. If the conditions of § 188 are satisfied except that $$\phi_{n}$$ tends steadily to a positive limit $$l$$, then the series $$\sum (-1)^{n}\phi_{n}$$ oscillates finitely.

6. Alteration of the sum of a conditionally convergent series by rearrangement of the terms. Let $$s$$ be the sum of the series $$1 – \frac{1}{2} + \frac{1}{3} – \frac{1}{4} + \dots$$, and $$s_{2n}$$ the sum of its first $$2n$$ terms, so that $$\lim s_{2n} = s$$.

Now consider the series $\begin{equation*} 1 + \tfrac{1}{3} – \tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{7} – \tfrac{1}{4} + \dots \tag{1} \end{equation*}$ in which two positive terms are followed by one negative term, and let $$t_{3n}$$ denote the sum of the first $$3n$$ terms. Then \begin{aligned} t_{3n} &= 1 + \frac{1}{3} + \dots + \frac{1}{4n-1} – \frac{1}{2} – \frac{1}{4} – \dots – \frac{1}{2n}\\ &= s_{2n} + \frac{1}{2n + 1} + \frac{1}{2n + 3} + \dots + \frac{1}{4n – 1}.\end{aligned}

Now $\lim \left[\frac{1}{2n + 1} – \frac{1}{2n + 2} + \frac{1}{2n + 3} – \dots + \frac{1}{4n – 1} – \frac{1}{4n}\right] = 0,$ since the sum of the terms inside the bracket is clearly less than $$n/(2n + 1)(2n + 2)$$; and $\lim \left(\frac{1}{2n + 2} + \frac{1}{2n + 4} + \dots + \frac{1}{4n}\right) = \tfrac{1}{2} \lim \frac{1}{n} \sum_{r=1}^{n} \frac{1}{1 + (r/n)} = \tfrac{1}{2} \int_{1}^{2} \frac{dx}{x},$ by § 156 and § 158. Hence $\lim t_{3n} = s + \tfrac{1}{2} \int_{1}^{2} \frac{dx}{x},$ and it follows that the sum of the series is not $$s$$, but the right-hand side of the last equation. Later on we shall give the actual values of the sums of the two series: see § 213 and Ch. IX, Misc. Ex. 19.

It can indeed be proved that a conditionally convergent series can always be so rearranged as to converge to any sum whatever, or to diverge to $$\infty$$ or to $$-\infty$$. For a proof we may refer to Bromwich’s Infinite Series, p. 68.

7. The series $1 + \frac{1}{\sqrt{3}} – \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{7}} – \frac{1}{\sqrt{4}} + \dots$ diverges to $$\infty$$. [Here $t_{3n} = s_{2n} + \frac{1}{\sqrt{2n + 1}} + \frac{1}{\sqrt{2n + 3}} + \dots + \frac{1}{\sqrt{4n – 1}} > s_{2n} + \frac{n}{\sqrt{4n – 1}},$ where $$s_{2n} = 1 – \dfrac{1}{\sqrt{2}} + \dots – \dfrac{1}{{\sqrt{2n}}}$$, which tends to a limit as $$n \to \infty$$.]