Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/244

Our results may then be stated as follows: Given the series of positive terms

$u_{1}+u_{2}+u_{3}+\cdots +u_{n}+u_{n+1}+\cdots$

find the limit

$\lim _{n\to \infty }\left({\frac {u_{n+1}}{u_{n}}}\right)=p$

I.	When $p<1$,[1] the series is convergent.
II.	When $p>1$, the series is divergent.
III.	When $p=1$, there is no test.

140. Alternating series. This is the name given to a series whose terms are alternately positive and negative. Such series occur frequently in practice and are of considerable importance.

If

$u_{1}-u_{2}+u_{3}-u_{4}+\cdots$

is an alternating series whose terms never increase in numerical value, and if

$\lim _{n\to \infty }u_{n}=0$

then the series is convergent.

Proof. The sum of $2n$ (an even number) terms may be written in the two forms

(A)	$S_{2n}=\left(u_{1}-u_{2}\right)+\left(u_{3}-u_{4}\right)+\left(u_{5}-u_{6}\right)+\cdots +\left(u_{2n-1}-u_{2n}\right)$ or
(B)	$S_{2n}=u_{1}-\left(u_{2}-u_{3}\right)-\left(u_{4}-u_{5}\right)-\cdots -u_{2n}$

Since each difference is positive (if it is not zero, and the assumption $\lim _{n\to \infty }u_{n}=0$ excludes equality of the terms of the series), series (A) shows that $S_{2n}$ is positive and increases with $n$, while series (B) shows that $S_{2n}$ is always less than $u_{1}$; therefore, by Theorem I, §136, $S_{2n}$ must approach a limit less than $u_{1}$ when $n$ increases, and the series is convergent.

Illustrative Example 4. Test the alternating series $1-{\tfrac {1}{2}}+{\tfrac {1}{3}}-{\tfrac {1}{4}}+\cdots$

Solution. Since each term is less in numerical value than the preceding one, and

$\lim _{n\to \infty }\left(u_{n}\right)=\lim _{n\to \infty }\left({\frac {1}{n}}\right)=0$

the series is convergent.

141. Absolute convergence. A series is said to be absolutely ^[2] or unconditionally convergent when the series formed from it by making all its terms positive is convergent. Other convergent series are said

1. It is not enough that $u_{n}+{\tfrac {1}{u_{n}}}$ becomes and remains less than unity for all values of $n$ but this test requires the limit of $u_{n}+{\tfrac {1}{u_{n}}}$ shall be less than unity. For instance, in the case of the harmonic series this ratio is always less than unity and yet the series diverges as we have seen. The limit, however, is not less than unity but equals unity.
2. The terms of the new series are the numerical (absolute) values of the terms of the given series.