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Problem Set 1 with Solution Key for First Year Interest Group Seminar | N 1, Assignments of Health sciences

Material Type: Assignment; Class: FIRST-YEAR INTEREST GROUP SMNR; Subject: Nursing; University: University of Texas - Austin; Term: Fall 2006;

Typology: Assignments

Pre 2010

Uploaded on 08/26/2009

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Download Problem Set 1 with Solution Key for First Year Interest Group Seminar | N 1 and more Assignments Health sciences in PDF only on Docsity!

M365C (Fall 2006) Due: Friday September 8, 2006 1

Instructions: Solve all the following problems. Your solutions will be graded for accuracy, thoroughness, and clarity.

1. Determine whether the sequence {

n

2

+ n − n }n∈N converges. If so, what is its limit? Prove all

your conclusions.

2. A sequence {xn}n∈N is said to be bounded if there exists b ∈ R such that |xn| ≤ b, for each n ∈ N.

Prove that every convergent sequence of real numbers is bounded.

3. a) Let X ⊂ R. We say that u ∈ R is an upper bound of X if x ≤ u for each x ∈ X. We say

s ∈ R is a least upper bound or a supremum of the set X, if s is an upper bound of X and

s ≤ u for every upper bound u of X. Prove that if a supremum x of X exists, then it is

unique. We use the notation sup X for the supremum of X.

b) Mimicking the above, define lower bounds and the greatest lower bound (or infimum) of

X ⊂ R. We use the notation inf X for the infimum of X. Prove the uniqueness of inf X if it

exists.

c) Suppose ∅ 6 = A ⊂ B ⊂ R. Prove that inf A ≥ inf B, and sup A ≤ sup B, assuming all the

quantities involved exist.

4. a) Prove the following:

i) Theorem Suppose {xn}n∈N, {yn}n∈N are sequences of real numbers such that limn→∞ xn =

+∞, and limn→∞ yn > 0. Then, limn→∞ xnyn = +∞.

ii) Theorem For a sequence {xn}n∈N of positive real numbers, limn→∞ xn = +∞ if and

only if limn→∞

xn

b) Prove, using the preceding theorems or otherwise, that

lim

n→∞

n

2

n + 1

= +∞, whereas lim

n→∞

n + 1

n

2

Remark: These examples show that the “limit theorems” established for sequences with finite

limits do NOT extend trivially to sequences with infinite limits. They also show that 0 · ∞

should remain undefined.

5. Prove that a x ∈ R is the limit of the sequence {xn}n∈N if and only if every subsequence of

{xn}n∈N has a further subsequence that converges to x.

M365C (Fall 2006) Due: Friday September 8, 2006 2

Solutions

  1. limn→∞

n 2

  • n − n

. Indeed, observe that

n 2

  • n − n −

n 2

  • n − n)(

n 2

  • n + n) √ n 2
  • n + n

n 2

  • n − n 2

√ n 2

  • n + n

2 n −

n 2

  • n − n √ n 2
  • n + n

n −

n 2

  • n

n +

n 2

  • n

n −

n 2

  • n

n +

n 2

  • n

n +

n 2

  • n

n +

n 2

  • n

n 2 − n 2 − n

(n +

n 2

  • n) 2

n

n 2

  • 2n

n 2

  • n + n 2
  • n

n

2 n 2

  • 2n

n 2

  • n + n

2 n + 2

n 2

  • n + 1

2 n

4 n

Hence, for any  > 0, choose any N >

. Then for any n > N , we have n > N >

, which implies the following:

4 N

4 n

n 2

  • n − n −

This proves that limn→∞

n 2

  • n − n

indeed exists and is equal to

  1. Let {xn}n∈N ⊂ R be a convergent sequence with limit L. Let  = 1. Then, there exists N ∈ N such that

|xn − L| <  = 1. Hence, L − 1 < xn < L + 1, for every n > N. Hence,

min{L − 1 , x 1 ,... , xN } < xn < max{L + 1, x 1 ,... , xN }, for every n ∈ N.

Hence {xn}n∈N is bounded.

  1. a) Suppose s and s ′ are both least upper bounds of the set X ⊂ R. Then, by the very definition of a least upper

bound, we have s ≤ s ′ and s ′ ≤ s. Hence, s = s ′ , which proves uniqueness of least upper bound (assuming it

exists).

b) Let X ⊂ R. We say that l ∈ R is a lower bound of X if l ≤ x for each x ∈ X. We say g ∈ R is a greatest

lower bound or an infimum of the set X, if g is a lower bound of X and l ≤ g for every lower bound l of X.

The infimum is unique, if it exists, because: Let g and g

′ be two infimums of the set X ⊂ R. Then, we must

have g ≤ g

′ and g

′ ≤ g; hence g = g

′ .

c) Since ∅ 6 = A ⊂ B, inf B is a lower bound of A and sup B is an upper bound of A. Hence, by definition of

infima and suprema, we immediately have the desired inequalities.