188. Alternating Series

1. The series $$\begin{array}{c}1–\frac{1}{2}+\frac{1}{3}–\frac{1}{4}+\dots ,1–\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}–\frac{1}{\sqrt{4}}+\dots ,\\ \sum \frac{(-1{)}^n}{(n+a)},\sum \frac{(-1{)}^n}{\sqrt{n+a}},\sum \frac{(-1{)}^n}{(\sqrt{n}+\sqrt{a})},\sum \frac{(-1{)}^n}{(\sqrt{n}+\sqrt{a}{)}^2},\end{array}$$ 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 $0<s\le 1$, and oscillatory if $s\le 0$.

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 $$\sum [\left\{\frac{(-1{)}^n}{\sqrt{n}+(-1{)}^n}–\frac{(-1{)}^n}{\sqrt{n}}\right\}+\frac{(-1{)}^n}{\sqrt{n}}]$$ 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 $limn{\chi }_n=1$. Hence the original series is divergent, although it is of the form ${\varphi }_2–{\varphi }_3+{\varphi }_4–\dots$, where ${\varphi }_n\to 0$. This example shows that the condition that ${\varphi }_n$ 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 ${\varphi }_n$ tends steadily to a positive limit $l$, then the series $\sum (-1{)}^n{\varphi }_n$ 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 $$\begin{array}{}\text{(1)}& 1+\frac{1}{3}–\frac{1}{2}+\frac{1}{5}+\frac{1}{7}–\frac{1}{4}+\dots \end{array}$$ in which two positive terms are followed by one negative term, and let $t_{3n}$ denote the sum of the first $3n$ terms. Then $$\begin{array}{rl}{t}_{3n}& =1+\frac{1}{3}+\cdots +\frac{1}{4n-1}–\frac{1}{2}–\frac{1}{4}–\cdots –\frac{1}{2n}\\ & =s_{2n}+\frac{1}{2n+1}+\frac{1}{2n+3}+\cdots +\frac{1}{4n–1}.\end{array}$$

Now $$lim[\frac{1}{2n+1}–\frac{1}{2n+2}+\frac{1}{2n+3}–\cdots +\frac{1}{4n–1}–\frac{1}{4n}]=0,$$ since the sum of the terms inside the bracket is clearly less than $n/(2n+1)(2n+2)$; and $$lim(\frac{1}{2n+2}+\frac{1}{2n+4}+\cdots +\frac{1}{4n})=\frac{1}{2}lim\frac{1}{n}\sum _{r=1}^n\frac{1}{1+(r/n)}=\frac{1}{2}{\int }_1^2\frac{dx}{x},$$ by § 156 and § 158. Hence $$lim{t}_{3n}=s+\frac{1}{2}{\int }_1^2\frac{dx}{x},$$ 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 $\mathrm{\infty }$ or to $-\mathrm{\infty }$. 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 $\mathrm{\infty }$. [Here $$t_{3n}=s_{2n}+\frac{1}{\sqrt{2n+1}}+\frac{1}{\sqrt{2n+3}}+\cdots +\frac{1}{\sqrt{4n–1}}>s_{2n}+\frac{n}{\sqrt{4n–1}},$$ where $s_{2n}=1–{\displaystyle \frac{1}{\sqrt{2}}}+\cdots –{\displaystyle \frac{1}{\sqrt{2n}}}$, which tends to a limit as $n\to \mathrm{\infty }$.]