Math 414
Professor Lieberman
May 6, 2003
HOMEWORK #13 SOLUTIONS
Section 6.7
11. FALSE. Make the substitution x = sin u, so dx = cos u du. Then
Z 1
Z π/2
x4
3u sin 2u sin 4u π/2 3π
√
dx =
sin4 u du =
−
+
=
.
8
4
32 0
16
1 − x2
0
0
33. TRUE. The upper and lower integrals are equal for any integrable function.
34. FALSE. Take
(
0
f (x) =
1
if x is rational,
if x is irrational
on the interval [0, 1] and A = 21 . Then
Z 1
Z 1
f =0≤A≤1=
f
0
0
but f is not integrable.
Section 7.1
1. (c) We can write this sum as
X 1
1
−
,
k+1 k+2
which is a telescoping series. Because lim 1/(k + 1) = 0, it follows that the series converges
k→∞
to 1/2.
√
(f) This series diverges because k + 1 does not tend to zero as k → ∞.
(j) We have
sin x
1
lim k sin = lim
= 1,
k→∞
k x→0 x
so the series diverges by the n-th term test.
3. (c) The repeating decimal is equal to
3.2 +
.015
1061
=
.
1 − .01
330
10. (a) First, we use induction to show that
n
X
bk = a1 − an+1 .
k=1
1
2
When n = 1, we have
X
bk = b1 = a1 − a2 .
k=1
Next, if the equation is true for n = m, then we have
m+1
X
k=1
bk =
m
X
bk + bm+1 = a1 − am+1 + (am+1 − am+2 ) = a1 − am+1 .
k=1
Now write (sn ) for the sequence of partial sums of bk , so
n
X
sn =
bk .
k=1
From our
P formula, we have sn = a1 − an+1 , so (an ) converges if and only (sn ) converges.
(b) If
bk converges, then we have
X
bk = lim sn = lim a1 − an+1 = a1 − lim an .
k
n→∞
n→∞
n→∞
14. Because an ≥ 0, it follows that
bn =
but
a1 + · · · + an
a1
≥ ,
n
n
n
X
a1
k=1
k
is just a1 times the harmonic series, which diverges. Therefore
parison test.
P
bk diverges by the com-