Answers & Hints for Selected Exercises 687
discontinuous. By Lebesgue's criterion (7.9.5), the set of points of discontinuity
off in [a, b] must have measure 0.
- Let 0 < € < 1/4, Qn[O, 1] = {rn: n EN}, and Vn EN, Jn= (rn - c/2n, rn + c/2n).
00
Then A = LJ Jn is open and bounded (why?). Let a = inf A, b = sup A , and let
n=l
P = {xo, x1, x2, · · · , Xn} be any partition of [a, b]. Since each [xi-l, xi] contains
n n
a rational number, S(f, P) = L Md~"i = L 6.i = b - a ~ 1, while$._(!, P) =
n i=l oo i=l oo
L mi6i = L {6.i: [xi-1, Xi]~ A} ::; L l(Jn) = L c/2n-l = 2c < 1/2.
i=l n=l n=l
.". l: f ~ 1 and l: f ::; 1/2, so f is not integrable on any interval containing A.
- (a) 1/99
(f) +oo
Chapter 8
EXERCISE SET 8.1
(b) 9876/9999 (c) 2/3 (d) 1/4
(g),(h),(j) diverge by general term test
(e) 15/2
(i) 0
- By Exercise 6.2.19, cosnx....,,.. 0, and sinx....,,.. 0 unless xis an integral multiple
of n, in which case sin nx = 0. Apply general term test.
00
- Let L bn be the result of altering or deleting a finite number of terms from
n=l
00
L an, and let am-1 be the last term altered or deleted. Then am = bm' for
n=l
00 00
some m' EN, and L an= L bn. Apply Ex. 4.
n=m n=m'
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- The nth partial sum of L (bn - bn+1) is Sn= b1 - bn+I· .".{Sn} converges
k=l
{::} {bn+1} converges, and n~oo lim Sn = b1 - n~oo lim bn.
- n=l E n2+~n+6 = n-->oo lim k=l £: [k!2 - k!3] = n-->oo lim [i -n!3] = i·
n n n
- Let So= 0 and Sn= L Xk· Then L Xk = L (Sk - Sk-1).
k=l k=l k=l
- (a) Suppose Lan converges. Let L bn be formed by inserting parentheses
in Lan. Then each partial sum of L bn is, after removing parentheses, a partial
sum of Lan, so the sequence of partial sums of L bn is a subsequence of the
sequence of partial sums of Lan. Thus, L bn = Lan.
