Calculus: Analytic Geometry and Calculus, with Vectors

12.2 Ratio test and integral test 601

etcetera. Thus the series

UN + uN+1 + uN+2 + UN+3 +

is dominated by the convergent series

IUNI + IuNlr + IUNIr2 + IUNlra+

and the conclusion follows. To prove the second part of the theorem,

choose a number R for which 1 < R < p. There is then an index N such

that I un,+1/unl > R when n? N and hence

(12.233) I uN+PI > I UNI R' (p = 0, 1, 2, .. .).

Since UN 0 0 and R > 1, this shows that lung -> co as n - and com-

pletes the proof of Theorem 12.23.

The remainder of the text of this section involvesa connection between

series and integrals which is both interesting and important. Every-

thing that we do can be easily understood and permanently remembered

Y

Y=f(x)

1

U'

2 3

3

M

n-1

n un-1

n+1 x

Figure 12.24

with the aid of Figure 12.24. We suppose that the terms of a series auk

and the values of a function f are related by the formula

uk = f(k) (k = 1,2,3, ...),

that f is positive and continuous and decreasing over the interval x? 1,

and that f(x) --+ 0 as x-+ oc. The left member of the formula

(12.241) k11 uk =f 1nf(x) dx + IT, + IT2I + I T,I

• ... + ITr_iI + un

is then the sum of the areas of the rectangles of heights u1, u2, , un

that stand upon the unit intervals with left (or left-hand) end points

at 1, 2, 3, , n. The first term of the right member is the area of the

region bounded by the graphs of the equations x = 1, x = n, y = 0, and

y = f(x). For each k, ITkI is the area of the triangular patch Tk bounded

by the graphs of x = k, x = k + 1, y = f(x), and y = Uk. Elementary

bookkeeping shows that the members of (12.241) are equal. The easiest