Calculus: Analytic Geometry and Calculus, with Vectors

(^600) Series
appropriate times. We begin by fumbling with the question whether
the series
(12.21)
converges when x = 0.99. We set U. = n2xn and obtain the first and
then the second of the formulas
un = n2(1 - 1--610 ')",
Since (1 - 1/n)n --* 1/e as n ---> cc, it is easy to reach the correct con-
clusion that uloo is of the order of magnitude of 10,000/e and that there
are several values of n for which un > 1000. This can make us suspect
that the series is not convergent, but it is still possible that the series may
converge to some relatively large number of the order of 106 or 1012.
Appreciation of usefulness of the ratio test can now be gained by noticing
that the simple calculation
(12.22) lim
n-->m
lim
n-. m
lim
n-
(n + 1)2xn+1
n2xn
1+
1
n IxI = IxI
12x + 22x2 + 32x3 + 42x4 + 52x6 + ...
uloo = 10,000(1 - 1L)100.
shows that the series (12.21) converges when IxI < 1 and hence when
x = 0.99.

Theorem 12.23 (ratio test) Let ul + u2 + 713 + be a series

of nonzero terms and suppose that

(12.231) lim

n- m

uu

1

P

In case p < 1, the series 'u,, is absolutely convergent and lim un = 0.

n-.

In case p > 1, the series diverges and lim Iunl = oo,

To prove the first part of the theorem, choose a number r for which

p < r < 1. There is then an index N such that

(12.232)

I

'U-+1 I < (n? N).

Giving n successively the values N, N + 1, N + 2, - yields the

formulas

IuN+1I < JuNI r

I uN+2I < IuN+,I r < iuNlr2

IuN+31 < IuN+2I r < I uNI r3

IUN+41 < I uN+3I r < JuNI r4