446. In the enunciation of the preceding article the student should notice the significance of the words "from and after a fixed term."

and by taking n sufficiently large we can make this ratio approximate to x as nearly as we please, and the ratio of each term to the preceding term will ultimately be x. Hence if x<1 the series is convergent.

Here we have a case of a convergent series in which the terms may increase up to a certain point and then begin to

terms do not begin to decrease until after the 100th term.

447. An infinite series in which all the terms are of the same sign is divergent if from and after some fixed term the ratio of each term to the preceding term is greater than unity, or equal to unity.

Let the fixed term be denoted by u1. If the ratio is equal to unity, each of the succeeding terms is equal to u1, and the sum of n terms is equal to nu1; hence the series is divergent.

If the ratio is greater than unity, each of the terms after the fixed term is greater than u1, and the sum of ʼn terms is greater than nu1; hence the series is divergent.

448. In the practical application of these tests, to avoid having to ascertain the particular term after which each term is greater or less than the preceding term, it is convenient to find the limit of when n is indefinitely increased; let this

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Ип-1

limit be denoted by X.

If λ<1, the series is convergent. [Art. 445.]

If λ>1, the series is divergent. [Art. 447.]

If λ=1, the series may be either convergent or divergent, and a further test will be required; for it may happen that

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Ип-1

<1 but continually approaching to 1 as its limit when n is in

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definitely increased. In this case we cannot name any finite quantity which is itself less than 1 and yet greater than λ. Hence the test of Art. 445 fails. If, however, ->1 but continually approaching to 1 as its limit, the series is divergent by Art. 447.

Un-1

We shall use "Lim "the limit of

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Ип-1

Un "" as an abbreviation of the words Un-1 when n is infinite."

Example 1. Find whether the series whose nth term is

(n+1)xn n2

Un

If x=1, then Lim =1, and a further test is required.

Un-1

If x 1 the series becomes 12+22+32+42+..., and is obviously divergent.

Example 3. In the series

a+(a+d)r+(a+2d)r2+...+(a+n−1. d) pn-1+....

a+(n−1)d
a+ (n-2)d

r=r;

thus if r<1 the series is convergent, and the sum is finite.

449. If there are two infinite series in each of which all the terms are positive, and if the ratio of the corresponding terms in the two series is always finite, the two series are both convergent, or both divergent.

and is therefore a finite quantity, L say;

··· U2+U2+Uz+...+Un= L (1'1+V2 + V z +

...

Hence if one series is finite in value, so is the other; if one series is infinite in value, so is the other; which proves the proposition.

450. The application of this principle is very important, for by means of it we can compare a given series with an auxiliary series whose convergency or divergency has been already established. The series discussed in the next article will frequently be found useful as an auxiliary series.

is always divergent except when p is positive and greater than 1.

CASE I. Let p>1.

2

2P

following eight terms together are less than ;;

Hence the series is less than

The first term is 1; the next two terms together are less 4 than -; the following four terms together are less than the 4p' 8

and so on.

8p

that is, less than a geometrical progression whose common ratio is less than 1, since p>1; hence the series is convergent.

2

2p

following eight terms together are greater than

on. Hence the series is greater than

4

8

or

the

2

[Art. 447.]

and is therefore divergent.

CASE III. Let p<1, or negative.

Each term is now greater than the corresponding term in Case II., therefore the series is divergent.

Hence the series is always divergent except in the case when p is positive and greater than unity.

Thus if un and vn denote the nth terms of the given series and the auxiliary series respectively, we have

hence Limun 1, and therefore the two series are both convergent

Vn

or both divergent. But the auxiliary series is divergent, therefore also the given series is divergent.

This completes the solution of Example 1. [Art. 448.]

452. To show that the expansion of (1+x)" by the Binomial Theorem is convergent when x<1.

Let ur, ur+1 represent the rth and (r+1)th terms of the expansion; then

When r>n+1, this ratio is negative; that is, from this point the terms are alternately positive and negative when x is positive, and always of the same sign when x is negative. Now when r is infinite, Lim "r+1=x numerically; therefore since

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x<1 the series is convergent if all the terms are of the same sign; and therefore still more is it convergent when some of the terms are positive and some negative. [Art. 444.]