The University of Sydney
School of Mathematics and Statistics
Solutions to Practice questions for Quiz 1
MATH2023: Analysis Semester 2, 2022
Web Page: http://www.maths.usyd.edu.au/u/UG/IM/MATH2023/
Lecturer: Milena Radnovic´
1. Let S1 = (1, 2]∩[
√
2,
√
5) and S2 = (1, 2]∪[
√
2,
√
5). Find the minimum, maximum,
infimum and supremum for each of these two sets.
Solution: Since S1 = [
√
2, 2], so minS1 = inf S1 =
√
2 and maxS1 = supS1 = 2.
On the other hand, S2 = (1,
√
5), so inf S2 = 1, supS2 =
√
5. S2 does not have
minimum and maximum.
2. For the sets S1 and S2 from the previous question, find how many integers, rational,
and irrational numbers each of them contains.
Solution: S1 and S2 contain only one integer: 2 ∈ S1, 2 ∈ S2. Each of the two
sets contains infinitely many rational and infinitely many irrational numbers.
3. Are there bounded sets A and B in R, such that
inf(A ∪ B) < inf A and inf(A ∪B) < inf B ?
Solution: No, since inf(A ∪B) = min{inf A, inf B}.
4. Calculate lim
n→∞
n3
n3 + n + 1
.
Solution: lim
n→∞
n3
n3 + n + 1
= lim
n→∞
1
1 + 1
n2
+ 1
n3
=
1
1 + 0 + 0
= 1
5. Calculate lim
n→∞
(−1)n
n3 + 2
.
Solution: The expression is the product of the bounded sequence (−1)n and the
sequence
1
n3 + 2
, which converges to zero. Thus, the limit equals 0.
6. Let xn =
5n2 + 5
6n2 + 5
. Is (xn) monotone, bounded, convergent?
Solution: lim
n→∞
xn =
5
6
, thus the sequence is convergent, which implies that it is
also bounded. Direct checking shows that the sequence is monotone decreasing.
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7. Let α be a given integer. Find lim
n→∞
nααn.
Solution:
lim
n→∞
nααn =
{
+∞, if α ≥ 1;
0, if α ∈ {0,−1}.
For α ≤ −2, the sequence splits into two divergent subsequences: lim
n→∞
(2n)αα2n =
+∞, lim
n→∞
(2n+ 1)αα2n+1 = −∞.
8. Calculate lim
n→∞
n · 5n + 2 · 10n
2n · 9n + 10n+1 .
Solution: lim
n→∞
n · 5n + 2 · 10n
2n · 9n + 10n+1 = limn→∞
n · (5/10)n + 2
2n · (9/10)n + 10 =
0 + 2
0 + 10
=
1
5
.
9. lim
n→∞
(
1 +
7
n
)n
Solution: lim
n→∞
(
1 +
7
n
)n
= lim
n→∞
(
1 +
1
n/7
)n
= lim
n→∞
((
1 +
1
n/7
)n/7)7
= e7.
10. Let the sequences (an), (bn), (cn), (dn), (fn) be given by:
an = n
(−1)n , bn =
1
2n
, cn =
1
2n+ 1
, dn =
1
n
, fn = (−1)n.
Which of the sequences (bn), (cn), (dn), (fn) are subsequences of (an)?
Solution: Only (cn).
11. What are the accumulation points of the sequences from the previous question?
Solution: (an) has subsequences a2n = 2n and a2n+1 =
1
2n+ 1
, thus 0 is its only
accumulation point.
(bn), (cn), (dn) all converge to zero, thus it is also their only accumulation point.
(fn) has accumulation points 1 and −1.
12. Which of the following series are convergent:
∞∑
n=1
1
n4 +
√
n
,
∞∑
n=1
1√
n
,
∞∑
n=1
2
n+
√
n
?
Solution: We have:
1
n4 +
√
n
<
1
n4
, so the first series is convergent by compari-
son the the convergent p-series for p = 4 > 1.
The second series is the p-series with p = 1/2 < 1, so it is divergent.
Since
2
n+
√
n
≥ 2
n+ n
=
1
n
, the third series is divergent by comparison to the
harmonic series.
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13. Which of the following series are convergent:
∞∑
n=1
1
(−10)n ,
∞∑
n=1
(−1)n,
∞∑
n=1
√
5n?
Solution: These are all geometric series. The only one whose quotient q satisfies
the condition |q| < 1 is the first one, so that is the only one which converges.
14. Calculate
∞∑
n=3
1
(−10)n .
Solution: This is a geometric series with quotient q = − 1
10
, so its sum is:
∞∑
n=3
1
(−10)n =
∞∑
n=0
1
(−10)n −
(
1− 1
10
+
1
102
)
= − 1
1100
.
15. Let sequence (an) be given by: a1 = 2, an+1 = 2+
√
an, for n ∈ N. Is that sequence
convergent? If it is, find its limit.
Solution: First, assume that the sequence is convergentfind the possible candi-
dates for the value of its limit α. Since α = lim
n→∞
an = lim
n→∞
an+1 = lim
n→∞
(2+
√
an) =
2 +
√
α, we get that
√
α is a solution of the quadratic equation x2 − x − 2 = 0.
That equation has solutions x1 = 2 and x2 = −1. Since
√
α > 0, we have
√
α = 2,
i.e. α = 4.
Now, we need to show that the sequence is indeed convergent. First, we notice
that clearly the sequence is positive and we will prove by induction that it is also
bounded from above, an < 4 for all n ∈ N:
• for n = 1, a1 = 2 < 4;
• if an < 4, then an+1 = 2 +
√
an < 2 +
√
4 = 4.
Second, we prove that the sequence is increasing: an < an+1 for all n ∈ N:
• for n = 1, a1 = 2 < 2 +
√
2 = a2;
• if an < an+1, then 2 +
√
an < 2 +
√
an+1, i.e. an+1 < an+2.
So, by the theorem on monotone and bounded sequences, (an) is convergent. The
only possible value for its limit is α = 4, so lim
n→∞
an = 4.
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