MATH 101, Section 212 (CSP)
Week 10: Marked Homework Solutions
2011 Mar 24
1. (a) [5] 1 - 13 + 15 - 1 7 + · · · = P1 n=1(-1)n-1 2n1-1 is an alternating series.
The series of its absolute values P1 n=1 (-1)n-1 2n1-1 = P1 n=1 2n1-1 = 1 + 13 + 1 5 + 17 + : : :
is divergent by the Integral Test: f(x) = 2x1-1 is continuous for x > 1 2, positive for x > 1 2,
and f0(x) = -(2x-2 1)2 < 0 so f(x) is decreasing for x > 12. Therefore f is continuous, positive
and decreasing on [1; 1). The improper integral
Z11 2x1- 1 dx = tlim !1 Z1t(2x - 1)-1 dx (substitution u = 2x - 1, du = 2 dx)
= lim
t!1
12
Z1 2t-1 u-1 du
=
12
lim
t!1
ln juj]2 1t-1
=
12
lim
t!1
ln(2t - 1) = 1
is divergent.
However, the series itself is convergent: bn = janj = 2n1-1 satisfies bn ≥ bn+1 for all
n = 1; 2; 3; : : : (2n1-1 ≥ 2(n+1) 1 -1 iff 2(n + 1) - 1 ≥ 2n - 1 iff 2n + 1 ≥ 2n - 1 iff 1 ≥ -1 which
is true; alternatively, show that f(x) = 2x1-1 is decreasing by showing that f0(x) is negative)
and limn!1 bn = 0 (limn!1 2n1-1 = limn!1 2-1(1 =n=n) = 0), so by the Alternating Series Test
P1 n=1(-1)n-1 2n1-1 is convergent.
The series is convergent but the series of its absolute values is divergent, so the series
1 - 1
3 + 15 - 1 7 + : : : is conditionally convergent.
(b) [3] P1 j=2 j((-lnj 1))j3 = 2 (ln12)3 - 3 (ln13)3 + 4 (ln14)3 - 5 (ln15)3 + : : : is an alternating series.
The series of its absolute values P1 j=2 j((-lnj 1))j3 = P1 j=2 j (lnj 1 )3 is convergent, by the Integral
Test (see Week 9: Marked Homework Assignment, question 1(b)).
Therefore the series P1 j=2 j((-lnj 1))j3 is absolutely convergent (which implies that it is convergent, so we don’t need the to apply the Alternating Series Test in this case).
(c) [3] limk!1 ak = limk!1(-1)k does not exist (fakg is the sequence f1; -1; 1; -1; 1; -1; : : :g
which does not get close to any unique finite number for all sufficiently large k) so by the
Test for Divergence, the series Pk=0(-1)k is divergent.
(d) [3] P1 ‘=0(-1)‘ 100 (2‘+1)! 2‘+1 = 100 - 100 3!3 + 100 5!5 - 100 7!7 + : : : is an alternating series. We use the
Ratio Test:
lim
‘!1
ja‘+1j
ja‘j = ‘lim !1
1002(‘+1)+1
(2(‘+1)+1)!
1002‘+1
(2‘+1)!
= lim
‘!1
1002‘+3
1002‘+1
(2‘ + 1)!
(2‘ + 3)!
= lim
‘!1
1002
(2‘ + 2)(2‘ + 3) = ‘lim !1
1002
‘2
2 + 2‘
2 + 3‘
= 0