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THE UNIVERSITY OF WARWICK
FIRST YEAR EXAMINATION: January 2011
Analysis I
Time Allowed: 1.5 hours
Read carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book.
Calculators are not needed and are not permitted in this examination.
ANSWER 3 QUESTIONS.
If you have answered more than the required 3 questions in this examination, you will only be given credit for your 3 best answers.
The numbers in the margin indicate approximately how many marks are available for each part of a question.
- a) Define what it means for a sequence (an) to be bounded. [2]
b) For what values of x is the following inequality valid:
|x + 2| ≤ 5.
[2]
c) Let the sequence (an) be defined according to a 1 = 1 and an =
3 an− 1 − a^2 n− 1 for n ≥ 2. Assume that (an) tends to a > 0. Find the numerical value of a. Motivate your answer. [2] d) Assume that (an) is a decreasing sequence. Which of the following statements are true: i) (an) has a limit. ii) (an) is eventually negative. iii) (an) tends to minus infinity. iv) (an) tends to zero. [2]
e) State, without proof, whether the following sequences tends to A) infinity, B) minus infinity, C) zero, D) a constant c 6 = 0 or or E) does not tend to either infinity, minus infinity or any real number.
1 Question 1 continued overleaf
Question 1 continued
i) (an) =
n^2 2 n
ii) (an) =
( (^2) n (^2) +sin(n) 5 n^2 +3n+
iii) (an) =
10 n 2
n
) 1 /n) , iv) (an) =
en^ cos(n)
. [4]
f) Prove that inf{a^2 n : n ∈ N} = 0 for any null sequence (an). [2]
g) An increasing sequence (an) satisfies an < 4 for every n ∈ N. State, without proof, whether (an) must converge. [1] h) State the Bolzano-Weierstrass Theorem and use it to decide whether the se- quence (cos(2n)) has a convergent subsequence. [3] i) Find a sequence (an) that satisfies |an+1 − an| < |an − an− 1 | for every n > 1 but (an) is not Cauchy. [2]
j) Prove that every Cauchy sequence is bounded. [3] k) Give an example of a diverging series
n=1 an^ such that^
an+ an →^ 1 as^ n^ → ∞.^ [2] l) Give an example of a converging series
n=1 an^ such that^
an+ an →^ 1 as^ n^ → ∞. [2]
m) Compute
n=
π 3 n^.^ [2] n) Is
n=
n+sin^2 n n^2 +cos n finite or infinite? Motivate your answer.^ [3] o) Is
n=
(2n)! n^2 n^ finite or infinite? Motivate your answer.^ [3]
- a) State and prove the Sandwich Theorem for sequences. [5] b) Prove Bernoulli’s inequality: If x > −1 and n is a natural number then
(1 + x)n^ ≥ 1 + nx.
[3]
c) Prove the following statement: If x > 0 then (x^1 /n) → 1. Hint: Consider the sequence (x^1 /n^ − 1). [7]
d) Prove that the sequence
(an) =
1 + cos^2 (n) 2 + cos(n)
) 1 /n)
tends to 1 as n tends to infinity. [5]
2 Question 3 continued overleaf
Question 4 continued
d) Consider the series
n=
n + 2 −
n
(i) State whether it converges or diverges. [1] (ii) Give a full proof of whether it converges or diverges. [3]
4 END
MATHEMATICS DEPARTMENT
FIRST YEAR UNDERGRADUATE EXAMS
Course Title: Analysis I
Model Solution No: 1
a) A sequence (an) is bounded if there exists a real number C such that |an| ≤ C for all n ∈ N. [2]
b) We have two cases. Case 1: If x ≥ −2. Then |x + 2| = x + 2 which implies that if x ≥ −2 then |x + 2| ≤ 5 if and only if x ≤ 3. So − 2 ≤ x ≤ 3 are all the solutions when x ≥ −2. Case 2: If x < −2. Then |x + 2| = −x − 2 which implies that if x < −2 then |x + 2| ≤ 5 if and only if −x ≤ 7. That is, the only solutions when x < −2 are − 7 ≤ x < −2. Putting the two cases together gives the answer |x + 2| < 5 if and only if − 7 ≤ x ≤ 3. [2]
c) Since an → a we can conclude that a =
3 a − a^2. Squaring both sides and rearranging terms implies that a(2a − 3) = 0 which has the solutions a = 0 and a = 3/2. But a > 0 by assumption so we can conclude that the limit is a = 3/2. [2]
d) None of the statements are true. Two marks for the right answer. One mark for three correct answers. No marks for two or less correct answers.
e) i) C) ii) D) iii) A) iv) E) One mark for each correct answer.
f) Firstly, 0 is a lower bound since a^2 n ≥ 0. Now, suppose > 0. Since (an) is null, there exists N ∈ N such that |an| <
for n > N. Hence, for such n, a^2 n = |an|^2 < . So is not a lower bound. [2]
g) Yes, it does since any increasing sequence that is bounded above converges by the completeness axiom. [1]
h) Bolzano-Weierstrass Any bounded sequence has a convergent subsequence. ( mark) Now, | cos(x)| ≤ 1 for every x ∈ R, so cos(2n) is bounded. (1 mark) It follows that there is a convergent subsequence from the Bolzano-Weierstrass theorem. ( mark) [3]
MATHEMATICS DEPARTMENT
FIRST YEAR UNDERGRADUATE EXAMS
Course Title: Analysis I
Model Solution No: 2
a) Sandwich Theorem for sequences: Suppose that (an) → l and (bn) → l. If an ≤ cn ≤ bn then (cn) → l. [2] Proof: Since (bn) → l there exist for each > 0 a natural number Nb, such that
l − < bn < l + (∗)
for all n > Nb,. Similarly, there exists for each > 0 a natural number Na, such that
l − < an < l + (∗∗)
for all n > Na,. Let > 0 and N = max(Na,, Nb,) then an ≤ cn ≤ bn, (∗) and (∗∗) implies that l − < cn < l + for all n > N. That is |cn − l| < for all n > N. We have proved that for each > 0 there exist an N such that |cn − l| < for all n > N. This is exactly the definition of (cn) → l. [3]
b) We will prove this by induction. The inequality is trivial for n = 1. Assume that the inequality holds for n = k. That is
(1 + x)k^ ≥ 1 + kx for all x > − 1. (∗)
For any x > −1 we can multiply both sides of (∗) by 1 + x > 0 without changing the inequality:
(1 + x)k+1^ = (1 + x)k^ · (1 + x) ≥ (1 + kx) · (1 + x) = 1 + (k + 1)x + kx^2 ≥ 1 + (1 + k)x.
The last inequality is valis since kx^2 > 0. We have thus shown that the Bernoulli inequality for n = k implies the Bernoulli inequality for n = k + 1. The inequality follows by induction. [3]
c) Let us assume that x ≥ 1. Then an = x^1 /n^ − 1 ≥ 0 since x^1 /n^ ≥ 1 for all x ≥ 1. By Bernoulli’s inequality, which is applicable since an ≥ 0 > −1, we may conclude that (1 + an)n^ ≥ 1 + nan. (∗) But (1 + an)n^ = (x^1 /n)n^ = x so we may conclude from (∗) that
x ≥ 1 + nan =⇒
x − 1 n
≥ an.
So 0 ≤ an ≤ x− n 1. By the product rule we can conclude that
(x− 1 n
→ 0 since 1 /n → 0 and (x − 1) is a constant sequence. It follows from the sandwich rule that (an) → 0. That is (x^1 /n^ − 1) → 0 which implies that (x^1 /n) → 1 by the sum rule. This proves the statement for all x ≥ 1. [5] If 0 < x < 1 then 1/x = y ≥ 1 which implies that (y^1 /n) → 1 by the previous reasoning. But x^1 /n^ =
y^1 /n^
where we used the quotient rule in the last step. It follows that (x^1 /n) → 1 if x > 0. [2]
d) Notice that, since − 1 ≤ cos(n) ≤ 1,
1 3
1 + cos^2 (n) 2 + cos(n)
Since x ≤ y is equivalent to xn^ ≤ yn^ for positive numbers we can conclude that ( 1 3
) 1 /n ≤ an ≤ 21 /n.
From c) we know that (1/3)^1 /n^ → 1 and 2^1 /n^ → 1 and therefore by the sandwich rule for sequences an → 1. [5]
(iv) The sequence must eventually involve only odd terms. They can be written
b 2 n− 1 =
3 − 4 n 4 n
which converges to -1 as n → ∞. [2]
d) A sequence (an) is Cauchy if, for any > 0, there exists N ∈ N such that |an−am| < for any n, m > N. [2]
e) First of all observe that, for n > m,
|an − am| = |an − an− 1 + an− 1 − an− 2 +... + am+1 − am|.
Hence, applying the triangle inequality to the right hand side,
|an − am| ≤
n∑− 1
k=m
|ak+1 − ak|.
(1 mark) Next use the fact that 0 ≤ |ak+1 − ak| < 2 −k^ to get
|an − am| ≤
∑^ n−^1
k=m
2 −k^ ≤
∑^ ∞
k=m
2 −k.
(1 mark) Now suppose > 0. Since
k=m 2
−k (^) → 0, there exists N ∈ N such that |
k=m 2
−k| < for every m > N. (1 mark)
It follows that, for n > m > N ,
|an − am| ≤
∑^ ∞
k=m
2 −k^ < .
If n = m then the difference |an − am| vanishes, and if m > n > N we just swap m, n in the argument above. (final mark) [4]
MATHEMATICS DEPARTMENT
FIRST YEAR UNDERGRADUATE EXAMS
Course Title: Analysis I
Model Solution No: 4
a) (i) If (an) is decreasing and null, then the series
(−1)n+1an is convergent. (ii) We have
∑^ ∞
k=N
(−1)kak = (aN − aN +1) + (aN +2 − aN +3) + ... ≥ 0 ,
since all brackets are positive. Also,
∑^ ∞
k=N
(−1)kak = aN − (aN +1 − aN +2) − (aN +3 − aN +4) − ... ≤ aN ,
since all brackets are positive. Then 0 ≤
k=N (−1)
kak ≤ aN.
(iii) Without the (−1)n, the sequence is decreasing and null. By similar consider- ations as above, we have
∣ ∣ ∣
∑^ N
k=
bk −
∑^ ∞
k=
bk
∣ ≤ |bN +1|.
It is then enough that |bN +1| ≤ 10 −^2 , which certainly holds for N ≥ 106.
b) Either prove the statements below, or give a counter-example.
(i) True statement. If (bN ) = (
∑N
k=1 ak) converges, then^ bN^ −^ bN^ −^1 →^ 0, i.e., aN → 0. (ii) Wrong statement. A counter-example is an = (^1) n. (iii) Wrong statement. A counter-example is an = (−1)
n √n.
c) (i) The series converges for |x| < 1 and for x = −1. (ii) The series converges absolutely for |x| < 1. (iii) The series diverges to +∞ for x ≥ 1. (iv) The series does not converge for |x| > 1 and for x = 1.
d) (i) The series
n=
n + 2 −
n
diverges.
(ii) Since (
n + 2 −
n)(
n + 2 +
n) = 2, we have
n + 2 −
n = √n+2+^2 √n ≥ √^1 n+2. The latter series is the shifted series of 1/
n, which diverges, as can be proved e.g. by integral bounds.
