A collection of limits

A Collection of Limits

March 28, 2011

Contents
1 Short theoretical introduction 2 Problems 3 Solutions 1 12 23

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Chapter 1

Short theoretical introduction
Consider a sequence of real numbers (an )n≥1 , and l ∈ R. We’ll say that l represents the limit of (an )n≥1 if any neighborhood of l contains all the terms of the sequence, starting from a certain index. We write this fact aslim an = l, n→∞ or an → l. We can rewrite the above definition into the following equivalence:
n→∞

lim an = l ⇔ (∀)V ∈ V(l), (∃)nV ∈ N∗ such that (∀)n ≥ nV ⇒ an ∈ V .

One can easily observe from this definition that if a sequence is constant then it’s limit is equal with the constant term. We’ll say that a sequence of real numbers (an )n≥1 is convergent if it has limit and lim an ∈ R, ordivergent if it doesn’t have a limit or if it has the limit n→∞ equal to ±∞. Theorem: If a sequence has limit, then this limit is unique. Proof: Consider a sequence (an )n≥1 ⊆ R which has two different limits l , l ∈ R. It follows that there exist two neighborhoods V ∈ V(l ) and V ∈ V(l ) such that V ∩ V = ∅. As an → l ⇒ (∃)n ∈ N∗ such that (∀)n ≥ n ⇒ an ∈ V . Also, since an → l ⇒ (∃)n ∈ N∗ such that(∀)n ≥ n ⇒ an ∈ V . Hence (∀)n ≥ max{n , n } we have an ∈ V ∩ V = ∅. Theorem: Consider a sequence of real numbers (an )n≥1 . Then we have: (i) lim an = l ∈ R ⇔ (∀)ε > 0, (∃)nε ∈ N∗ such that (∀)n ≥ nε ⇒ |an − l| < ε.
n→∞

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A Collection of Limits

(ii) lim an = ∞ ⇔ (∀)ε > 0, (∃)nε ∈ N∗ such that (∀)n ≥ nε ⇒ an > ε.
n→∞

(iii) lim an = −∞ ⇔ (∀)ε > 0, (∃)nε ∈ N∗ such that (∀)n ≥ nε ⇒an < −ε
n→∞

Theorem: Let (an )n≥1 a sequence of real numbers. 1. If lim an = l, then any subsequence of (an )n≥1 has the limit equal to l.
n→∞

2. If there exist two subsequences of (an )n≥1 with different limits, then the sequence (an )n≥1 is divergent. 3. If there exist two subsequences of (an )n≥1 which cover it and have a common limit, then lim an = l.
n→∞

Definition: A sequence (xn)n≥1 is a Cauchy sequence if (∀)ε > 0, (∃)nε ∈ N such that |xn+p − xn | < ε, (∀)n ≥ nε , (∀)p ∈ N. Theorem: A sequence of real numbers is convergent if and only if it is a Cauchy sequence. Theorem: Any increasing and unbounded sequence has the limit ∞. Theorem: Any increasing and bounded sequence converge to the upper bound of the sequence. Theorem: Any convergent sequence is bounded. Theorem(Cesarolemma): Any bounded sequence of real numbers contains at least one convergent subsequence. Theorem(Weierstrass theorem): Any monotonic and bounded sequence is convergent. Theorem: Any monotonic sequence of real numbers has limit. Theorem: Consider two convergent sequences (an )n≥1 and (bn )n≥1 such that an ≤ bn , (∀)n ∈ N∗ . Then we have lim an ≤ lim bn .
n→∞ n→∞

Theorem: Consider a convergentsequence (an )n≥1 and a real number a such that an ≤ a, (∀)n ∈ N∗ . Then lim an ≤ a.
n→∞

Theorem: Consider a convergent sequence (an )n≥1 such that lim an = a.
n→∞

Them lim |an | = |a|.
n→∞

Short teoretical introduction

3

Theorem: Consider two sequences of real numbers (an )n≥1 and (bn )n≥1 such that an ≤ bn , (∀)n ∈ N∗ . Then: 1. If lim an = ∞ it follows that lim bn = ∞.
n→∞n→∞

2. If lim bn = −∞ it follows that lim an = −∞.
n→∞ n→∞

Limit operations: Consider two sequences an and bn which have limit. Then we have: 1. lim (an + bn ) = lim an + lim bn (except the case (∞, −∞)).
n→∞ n→∞ n→∞

2. lim (an · bn ) = lim an · lim bn (except the cases (0, ±∞)).
n→∞ n→∞ n→∞

3. lim

lim an an = n→∞ (except the cases (0, 0), (±∞, ±∞)). n→∞ bn lim bn
n→∞ lim bn

4.lim abn = ( lim an )n→∞ n
n→∞ n→∞

(except the cases (1, ±∞), (∞, 0), (0, 0)).

5. lim (logan bn ) = log lim a ( lim bn ). n n→∞ n→∞
n→∞

Trivial consequences: 1. lim (an − bn ) = lim an − lim bn ;
n→∞ n→∞ n→∞

2. lim (λan ) = λ lim an (λ ∈ R);
n→∞ n→∞

3. lim

√ k

n→∞

an =

k

n→∞

lim an (k ∈ N);

Theorem (Squeeze theorem): Let (an )n≥1 , (bn )n≥1 , (cn )n≥1...