series found by summation is s 1 = .1 , s 2 = .11 , s 3 = .111 , s 4 = .1111 , etc. The limit of the terms of this series is 1 9 ; this is easy to see by simple division, for 1 9 = .1 + 1 90 = .11 + 1 900 = .111 + 1 9000 = etc. Hence, if 3 17 is given (the k of the definition), .1 and all succeeding terms differ from 1 9 by less than 3 17 ; if 1 1000 is given (another choice for the k of the definition), .111 and all succeeding terms differ from 1 9 by less than 1 1000 ; and so on, whatever choice for k be made.

It is evident that nothing that has been said gives the slightest idea as to how the "sum to infinity" of a series is to be found. We have merely stated the conditions which such a number is to satisfy. Indeed, a general method for finding in all cases the sum to infinity of a series is intrinsically out of the question, for the simple reason that such a "sum," as here defined, does not always exist. Series which possess a sum to

infinity are called convergent, and those which do not possess a sum to infinity are called divergent.

An obvious example of a divergent series is $1$ , $2$ , $3$ , , $n$ , i.e. the series of integers in their order of magnitude. For whatever number $l$ you try to take as its sum to infinity, and whatever standard of approximation $k$ you choose, by taking enough terms of the series you can always make their sum differ from $l$ by more than $k$ . Again, another example of a divergent series is $1$ , $1$ , $1$ , etc., i.e. the series of which each term is equal to $1$ . Then the sum of $n$ terms is $n$ , and this sum grows without limit as $n$ increases. Again, another example of a divergent series is $1$ , $- 1$ , $1$ , $- 1$ , $1$ , $- 1$ , etc., i.e. the series in which the terms are alternately $1$ and $- 1$ . The sum of an odd number of terms is $1$ , and of an even number of terms is $0$ . Hence the terms of the series $s_{1}$ , $s_{2}$ , $s_{3}$ , , $s_{n}$ , do not approximate to a limit, although they do not increase without limit.

It is tempting to suppose that the condition for $u_{1}$ , $u_{2}$ , , $u_{n}$ , to have a sum to infinity is that $u_{n}$ should decrease indefinitely as $n$ increases. Mathematics would be a much easier science than it is, if this were the case. Unfortunately the supposition is not true.

For example the series $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots, \frac{1}{n}, \dots$ is divergent. It is easy to see that this is the case; for consider the sum of $n$ terms

beginning at the $(n + 1)$ th term. These $n$ terms are $\frac{1}{n + 1}$ , $\frac{1}{n + 2}$ , $\frac{1}{n + 3}$ , , $\frac{1}{2 n}$ : there are $n$ of them and $\frac{1}{2 n}$ is the least among them. Hence their sum is greater than $n$ times $\frac{1}{2 n}$ , i.e. is greater than $\frac{1}{2}$ . Now, without altering the sum to infinity, if it exist, we can add together neighbouring terms, and obtain the series $1, \frac{1}{2}, \frac{1}{3} + \frac{1}{4}, \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}, etc.,$ that is, by what has been said above, a series whose terms after the $2$ nd are greater than those of the series, $1, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, etc.,$ where all the terms after the first are equal. But this series is divergent. Hence the original series is divergent.Cf. Note C, noteC.204

This question of divergency shows how careful we must be in arguing from the properties

of the sum of a finite number of terms to that of the sum of an infinite series. For the most elementary property of a finite number of terms is that of course they possess a sum: but even this fundamental property is not necessarily possessed by an infinite series. This caution merely states that we must not be misled by the suggestion of the technical term "sum of an infinite series." It is usual to indicate the sum of the infinite series $u_{1}, u_{2}, u_{3}, \dots, u_{n}, \dots$ by $u_{1} + u_{2} + u_{3} + \dots + u_{n} + \dots .$