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Math 1352-11 — WW07 Solutions
November 15, 2008
Assigned problems: 8.1 – 12, 16, 26, 8.2 – 4, 6, 12, 28, ww 8, 8.3 – 4, 12, 20, 36
Always read through the solution sets even if your answer was correct. Note that like many of the integrals in this course, there is frequently more than one way to determine convergence or divergence of a series. Your solution may be correct even if you used a different method than what I use here.
- (8.1 #12)
nlim→∞
5 n + 8 n
= (^) nlim→∞
5 + 8/n 1
Sequence converges to 5.
- (8.1 #16)
nlim→∞
8 n^2 + 800n + 5000 2 n^2 − 1000 n + 2 =^ nlim→∞
16 n + 800 4 n − 1000 (using L’hopital’s rule) = (^) nlim→∞^16 4
(L’hop. again) = 4
Sequence converges to 4.
- (8.1 #26)
nlim→∞
n
)n
This is similar to a limit you should remember from Calc I:
nlim→∞
1 +^1
n
)n = e
(If you do not remember this limit, review section 2.4 of the textbook. I will assume that you know this limit.) Here, we need to do a substitution to turn the given limit into one involving the limit we know. Let u = n/3. This means that n = 3u and 3/n = 1/u. Note also that u → ∞ as n → ∞. So we can rewrite our limit as:
nlim→∞
1 +^3
n
)n = (^) ulim→∞
1 +^1
u
) 3 u
= (^) ulim→∞
[(
u
)u] 3
[
u^ lim→∞
1 +^1
u
)u] 3 = e^3
Sequence converges to e^3.
Copyright 2008, Victoria Howle and Department of Mathematics & Statistics, Texas Tech University. All rights reserved. No part of this document may be reproduced, redistributed, or transmitted
∑^ ∞
k=
)k
Geometric series with a = 1 and r = − 4 /5. Converges since |r| = 4/ 5 < 1. Converges to:
a 1 − r
=^5
Series converges to 5/9.
- (8.2 #6)
∑^ ∞
k=
(−3)k^ =
∑^ ∞
k=
)k
Geometric series with a = 2 and r = − 13. Converges since |r| = 1/ 3 < 1. Converges to:
a 1 − r =^
Series converges to 3/2.
- (8.2 #12)
∑^ ∞
k=
3 k 4 k+^
∑^ ∞
k=
3 k 424 k^
∑^ ∞
k=
)k
Geometric series with r = 3/4. Converges since |r| = 3/ 4 < 1. Converges to:
a 1 − r
3 43 1 − (^34)
(Note that ”a” is the first term in the series, which is when k = 1 in this case.)
Series converges to 3/16.
Copyright 2008, Victoria Howle and Department of Mathematics & Statistics, Texas Tech University. All rights reserved. No part of this document may be reproduced, redistributed, or transmitted
∑^ ∞
n=
[ln(2(n + 1)) − ln(2n)]
I’ll try the divergence test first. The sequence diverges if limn→∞ an 6 = 0.
nlim→∞ [ln(2(n^ + 1))^ −^ ln(2n)]^ =^ nlim→∞ ln
(2(n + 1)) (2n)
= ln
n^ lim→∞
2 n + 2 2 n
= ln
n^ lim→∞
2 + 2/n 2
= ln(1) = 0
The limit is 0, so the divergence test is inconclusive (may converge or diverge). We’ll try looking at the nth partial sum.
Sn = [ln(4) − ln(2)] + [ln(6) − ln(4)] + [ln(8) − ln(6)] + · · · + [ln(2n) − ln(2n − 2)] + [ln(2n + 2) − ln(2n)] = [ln(4) − ln(2)] + [ln(6) − (^) ln(4)] + [ ln(8) − (^) ln(6)] + · · · + [ln(2n) − (^) ln(2n − 2)] + [ln(2n + 2) − (^) ln(2n)] = − ln(2) + ln(2n + 2)
S = (^) nlim→∞ Sn = (^) nlim→∞ [− ln(2) + ln(2n + 2)] = ∞
The limit goes to infinity, so the series diverges.
- (8.3 #4)
∑^ ∞
k=
k
∑^ ∞
k=
k^1 /^2
Divergent p-series since p = 12 ≤ 1.
Copyright 2008, Victoria Howle and Department of Mathematics & Statistics, Texas Tech University. All rights reserved. No part of this document may be reproduced, redistributed, or transmitted
1
xe−x
2 dx = lim N →∞
∫ N
1
xe−x
2 dx
(Let u = x^2 , du = 2x dx)
= lim N →∞
∫ (^) x=N
x=
e−u^ du
= lim N →∞
e−u
x=N
x=
= (^) Nlim →∞ −
e−x
2
N
1 = lim N →∞
e−N^
2
e−^1
= lim N →∞
2 eN^2
2 e
2 e So the improper integral converges. Next we need to determine if the integral test applies to the given series. To apply, the integrand must be positive, continuous, and decreasing. It is positive since xe−x^2 is positive for all x > 0. It is a rational function of continuous functions, so continuous. We just need to show that it is decreasing, i.e., that f ′(x) < 0 if f (x) = xe−x
2 .
f (x) = xe−x
2
f ′(x) = e−x
2
e−x
(− 2 x)
=
ex^2
2 x^2 ex^2 = 1 −^2 x
2 ex^2
which is negative for all x greater than or equal to 1. Therefore, the function is decreasing for x ≥ 1. The function is positive, continuous, and decreasing, so the integral test applies and the improper integral and the series converge or diverge together. The integral converges, therefore the series converges.
Copyright 2008, Victoria Howle and Department of Mathematics & Statistics, Texas Tech University. All rights reserved. No part of this document may be reproduced, redistributed, or transmitted
