Calculus: Analytic Geometry and Calculus, with Vectors

12.2 Ratio test and integral test 599

Let e > 0. Choose an index P such that

(4) Ixn - LI < e/2 (n > P)

Then, when both m and n exceed P,

(5) Ixm - x,I = I (x, - L) - (x,, - L) I < Ixm - LI + Ixn - LI

<e/2+e/2=e,

and the conclusion follows. It is also true that each Cauchy sequence is convergent

Proofs of this fact are much more difficult because they must, in one way or

another, use completeness of the real-number system to produce the number to

which the sequence converges. One lively proof starts with the choice of an

increasing sequence Pi, P2, of integers such that, for each k = 1, 2, 3, .,

(6) IXm - XnI < 2 *n > Pk).

Then

1

(7) I XPt+i - XPjj <2k (k = 1, 2, 3, ...).

It follows from the comparison test for convergence of series that the series

(8) XPt + (XP, - xP) + (XP7 - XP') + (XP4 - xP,) +...

is convergent, say to L. Since the sum of n terms of this series is xp,,, it follows
that

(9) lim xP = L.

This shows that a subsequence of the given sequence converges to L. To prove

that the whole sequence converges to L, let e > 0. Choose a positive number N

such that

(10) Ixn - x,I < e/2

and

(11) Ixp - LI < e/2

(m,n > N)

(n > N).

Then, when n > N, we can use the above inequalities and the fact that P. _!: n

to obtain

(12) Ixn - LI = I (x. - XP,.) + (xP,. - L)I

5 Ixn- xP I + Ixp,. - LI<e/2 + e/2 = e,

and our result is established.

12.2 Ratio test and integral test

theorems about convergence of series. These theorems, like hammers

and saws and other tools in carpenter shops, have their usefulnesses and

we can cultivate abilities to make effective use of appropriate ones at