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Math253(CalculusIII), Winter 2019
Practice Final Solutions
1. (28 pts) Does the sequence converge or diverge? WHY? If it converges, what is the
limit?
3n3 +16n
(a) an = cos(16n3 +12n2 +3)
Solution: By examination of leading terms, lim 3n3+16n =lim 3n3 = 3 . Since
16n3+12n2+3 16n3 16
cosisacontinuousfunction,wegetlima = cos( 3 ),andthesequenceconverges.
n 16
(b) b = (1;0; 1;0;0; 1;0;0;0; 1;0;0;0;0;:::)
2 3 4 5
Solution: This sequence converges to zero. There are many reasonable explana-
tions, and most of them use words! Something naive like “It switches (after vary-
ingamountsoftime)betweentwosequenceswhichbothconvergetozero”would
befine. I would probably accept anything reasonable and not erroneous.
It does NOT “alternate” between the sequence 1 and the sequence 0. The word
n
“alternate” really implies taking turns.
One can NOT use the following argument: we have 0 ≤ b ≤ 1 so the squeeze
n n
theorem implies that limb = 0. Why not? (Answer: because b = 1 > 1.) One
n 6 4 6
1
coulduse0≤b ≤ √ butthatistricky!
n n
n2
(c) cn = en
Solution: By extension to a function and L’Hopital’s rule we have
n2 2n 2
lim en = lim en = lim en = 0;
since 2 is bounded and en increases to infinity.
(d) dn = n−20 +(−1)n
n+3
Solution: This diverges. By examining leading terms, n−20 converges to 1. How-
n+3
ever, (−1)n diverges, and by the 2/3 rule, the sum of convergent and divergent is
divergent.
2
2. (21 pts) Does the series converge or diverge? WHY?
∞ 2 n
(a) X n 5
2n+4
n=3 3
2 n+1 2n+4 2
Solution: We do the ratio test. lim (n+1) 5 3 = lim (n+1) 5 = 5. Since
2 n 2(n+1)+4 2 2
n 5 3 n 3 9
this is less than 1, the sequence converges.
∞ 5
Xsin(n )
(b) √
n=1 n n
5 5 P |sin(n5)|
Solution: We have −1 ≤ sin(n ) ≤ 1 so |sin(n )| ≤ 1. Now n=1 n√n con-
P |sin(n5)| P 1
verges by the comparison test, since n=1 n√n ≤ n=1n√n whichconverges
bythep-test, p = 1:5. So the original sequence converges by the absolute conver-
gencetest.
∞ 3
(c) X(−1)nn3+1
n=0 n +2
Solution: Since lim n3+1 = 1, the sequence (−1)nn3+1 diverges (it alternates be-
n3+2 n3+2
tween a sequence converging to 1 and a sequence converging to −1, so it never
stays close to either one). By the divergence test, this series diverges.
3
3. (21 pts) Does the series converge or diverge? WHY?
∞
(a) X n+2
13n2 +12
n=15
Solution: Looking at leading terms, we expect this to behave like P 1 , which
13n
diverges. To prove it diverges, we use the comparison test:
n+2 ≥1· 1
13n2 +12 2 13n
for large enough n, and P 1 · 1 diverges by the p-test.
2 13n
(b) 1− 1 − 1 + 1 − 1 − 1 − 1 + 1 −:::
2 4 8 16 32 64 128
Solution: This converges because it absolutely converges. The sum of the absolute
values is P∞ 1 , which is a convergent geometric series with r = 1.
n=0 n
2 2
∞
(c) X 1 (Hint: whatisthederivativeofln(ln(t))?)
n=2 nlnn
Solution: The derivative of ln(ln(t)) is 1 . Since limt→∞ln(ln(t)) = ∞, we ex-
tlnt
pect this series to diverge. In fact, by the integral test P∞ 1 ≥R∞ 1 dt =
n=2 nlnn 2 tlnt
ln(ln(t)) |∞= ∞.
2
4. (14 pts) If the series converges, find the sum, and justify your answer. If it diverges,
explain why.
∞
(a) X6(11)n
n=5 10
Solution: This is a geometric series with a = 6(11)5 and r = 11. Since r > 1, the
10 10
series diverges.
(b) 3+2+ 4 + 8 + 16 +:::
3 9 27
Solution: This is a geometric series with a = 3 and r = 2. Since r < 1, the series
a 3 3
converges to 1−r = 1−2 = 9.
3
4
∞
5. (16 pts) Consider the convergent series X 1 .
n4
n=1
(a) Find an upper bound for the difference between 1 + 1 + 1 and the sum of the
4 4
2 3
series.
Solution: By the integral test,
∞ Z ∞
X1 ≤ 1 dt = −1 |∞= 1 = 1 :
4 4 3 3 3
n=4 n 3 t 3t 3·3 81
(b) How many terms of the series must one add to approximate the sum to within
1 ?
300
Solution: By the same argument,
∞ Z ∞
X 1 ≤ 1 dt = 1 :
n4 t4 3N3
n=N+1 N
When N = 5 we have N3 > 100 so 3N3 > 300 so 1 < 1 . Five terms is
3N3 300
sufficient.
It’s also fine to solve for N and say that N is greater than the cube root of 100.
6. (20 pts) Consider the series − 3 + 3 − 3 + 3 −:::.
11 14 17 20
(a) Find an explicit formula for this series.
Solution: This is P∞ (−1)n 3 . (One could also do P∞ (−1)n−1 3 .)
n=1 8+3n n=0 11+3n
(b) Is the series convergent? WHY?
Solution: Yes, by the Alternating Series Test, because lim 3 =0.
8+3n
(c) Is the series absolutely convergent? WHY?
Solution: No, by the comparison test, because P 3 ≥P1·1 whichdiverges
3n+8 2 n
byp-test, p = 1.
(d) Howmanytermsshouldoneaddtoapproximatethesumtowithin:01?
Solution: The error is bounded by the absolute value of the next term. We have
3 < 1 ifandonlyif300 < 11+3(N+1)=14+N ifandonlyifN > 286.
11+3(N+1) 100 3
(So N = 96 terms will do.)
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