Page 112 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 5] INTEGRALS 103

1 1 1
5.4. Evaluate lim þ þ þ .
n!1 n þ 1 n þ 2 n þ n
The required limit can be written
n
1 1 1 1 1 X 1
lim ¼ lim
n!1 n 1 þ 1=n þ 1 þ 2=n þ þ 1 þ n=n n!1 n 1 þ k=n
k¼1
ð 1 dx
1
¼ lnð1 þ xÞj 0 ¼ ln 2
0 1 þ x
¼
using Problem 5.2 and the fundamental theorem of the calculus.

1 t 2t ðn 1Þt 1 cos t
5.5. Prove that lim sin þ sin þ þ sin .
n!1 n n n n ¼ t
Let a ¼ 0; b ¼ t; f ðxÞ¼ sin x in Problem 1. Then
n
t X kt ð t
lim sin sin xdx ¼ 1 cos t
n!1 n n ¼
k¼1 0
and so
n 1
1 X kt 1 cos t
lim sin
n!1 n n ¼ t
k¼1
sin t
using the fact that lim ¼ 0.
n!1 n

MEASURE ZERO

5.6. Prove that a countable point set has measure zero.
Let the point set be denoted by x 1 ; x 2 ; x 3 ; x 4 ; ... and suppose that intervals of lengths less than
=2; =4; =8; =16; ... respectively enclose the points, where is any positive number. Then the sum of
1
the lengths of the intervals is less than =2 þ =4 þ =8 þ ¼ (let a ¼ =2 and r ¼ in Problem 2.25(a)of
2
Chapter 2), showing that the set has measure zero.

PROPERTIES OF DEFINITE INTEGRALS

ð b ð b

f ðxÞ dx @ j f ðxÞj dx if a < b.

5.7. Prove that
a a
By absolute value property 2, Page 3,

n n n

X X X
f ð k Þ x k @

j f ð k Þ x k j¼ j f ð k Þj x k

k¼1 k¼1 k¼1
Taking the limit as n !1 and each x k ! 0, we have the required result.
ð 2 sin nx
5.8. Prove that lim dx ¼ 0.
2
0 x þ n 2
n!1
2 2 ð 2
ð ð
sin nx sin nx dx 2
dx @ dx @ ¼

0 x þ n 0 x þ n 0 n
2 2 2 2 2 n 2
2
ð
sin nx
Then lim dx ¼ 0, and so the required result follows.

0 x þ n
2 2
n!1