Exercise 126.

Find the limiting values of:

1.

2.

(4 x2 — 3)(1 − 2 x) when x becomes infinitesimal.
7x-6x+4

(x2 − 5)(x2+7) when x becomes infinite.

x+35

12. If x approaches a as a limit, and n is a positive integer, show that the limit of 2" is a".

13. If x approaches a as a limit, and a is not 0, show that the limit of x" is a", where n is a negative integer.

CHAPTER XXXIV.

SERIES.

435. Convergency of Series. For an infinite series to be convergent (§ 325) it is necessary and sufficient that the sum of all the terms after the nth, as n is indefinitely increased, should approach 0 as a limit.

Although each of the terms after the nth may approach O as a limit, their sum may not approach O as a limit. Thus, take the harmonical series,

Each term after the nth approaches O as n increases.
The sum of n terms after the nth term is

Now, the first term is 1, the second term is, the sum of the next two terms is greater than, the sum of the succeeding four terms is greater than; and so on. So that, by increasing n indefinitely, the sum will become greater than any finite multiple of §. Therefore, the series is divergent.

To determine whether the following series is convergent:

But as n increases indefinitely, this last expression approaches O as a limit. Hence, the series is convergent.

436. Test for Convergency of a Series. If the terms of an infinite series are all positive, and the limit of the nth term is 0, then if the limit of the ratio of the (n+1)th term to the nth term, as n is indefinitely increased, is less than 1, the series is convergent.

.....

Let u1, U2, uz,
Un, Un+1, Un+2, be an infinite series.
Let r represent the limit of the ratio Un+1 as n increases

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indefinitely, and suppose r to be positive and less than 1. Let k be some fixed number between r and 1, and take k so near 1 that Un+1 Un+2 shall each be <k.

Un

Un+1

But, by hypothesis, u, approaches 0 as a limit as n is indefinitely increased. Hence, the series is convergent. Similarly, when r is negative, and between 0 and 1.

.....

k

and this approaches 0 as a limit as n is indefinitely in

creased; moreover, the nth term,

Hence, the series is convergent.

If r>1, there must be in the series some term from which the succeeding term is greater than the next preceding term; so that the remaining terms will form an increasing series, and therefore the series is not convergent.

If r±1, this value gives no information as to whether the series is convergent or not; and in such cases other tests must be applied.

If r<1, but approaches 1, or 1, as a limit, then no fixed value k can be found which will always lie between r and ± 1, and other tests of convergency must be applied. Thus, in the infinite series

r, the ratio of the (n + 1)th term to the nth term, is

which approaches 1 as a limit as n increases.

Suppose m positive and greater than 1; then the first term of the

series is 1. The sum of the next two terms is less than

2

The sum

2m

4

of the next four terms is less than

The sum of the next eight

4m

and so on. Hence, the sum of the series is less

8m

which is evidently convergent when m is positive and greater than 1.

If m is positive and equal to 1, the given series becomes

which is the harmonical series shown in ? 435 to be divergent.

If m is negative, or less than 1, each term of the series is then greater than the corresponding term in the harmonical series, and hence the series is divergent.

437. Special Case. If the terms of an infinite series are alternately positive and negative; if, also, the terms continually decrease, and the limit of the nth term is zero, then the series is convergent.

The sum of the terms after the nth term is

± [Un+1 − (Un+2 —— Un+3) — (Un+4 — Un+5) — ·····],

which may be written

±[Un+1 — Un+2+(Un+3 − Un+4) + (Un+5 — Un+6) + ·····].

Since the terms are continually diminishing, each of the groups in either form of expression is positive, and therefore the absolute value of the required sum is seen, from the first form of expression, to be less than u+1; and from the second form of expression, to be greater than un+1-Un+2. But both Un+1 and Un+2 approach zero as n increases indefinitely; therefore the sum of the series after the nth term approaches zero, and the series is convergent.

In finding the sum of an infinite decreasing series of which the terms are alternately positive and negative, if we stop at any term, the error will be less than the next succeeding term.