Convergence of Sequences

A natural inquiry abount a sequence {a_n} is whether the terms a_n come close to any real number when n is extremely large. This is what is known as the convergence of a sequence.

Definition 2.2.1. (Limit of a sequence) A sequence {a_n} is said to converge to a real number ℓ, called the limit of {a_n}, if for every ε > there exists a natural number n₀ such that:

If the sequence {a_n} converges to the limit ℓ, then we write

note that the inequality |a_n − ℓ| < ε corresponds to the chain of inequalities:

from this relation is clear that: every convergent sequence is as well bounded.

Example 2.2.2. Let a_n = 1/n, n = 1, 2, ... . Then the sequence converges to 0. For let ε > 0. Choose N to be the next integer after 1/ε (we use the Archimedean property, if r,ε are positive real numbers, then there is a positive integer n sucht that nε > r). If n > N then

proving the claim.  ■

Example 2.2.3. The sequence a_n = (n² + 1)/(n² + 2n − 1) has limit 1; indeed

for n > 2/ε. It is not generally necessary to find the least n₀ such that the previous relation is satisfied.  ■

In some cases ∞ or −∞ can be regarded as “limits” of a sequence {a_n}. We say that {a_n} tends to ∞ or diverges to ∞ if for every α ∈ ℝ, there is n₀ ∈ ℕ such that a_n > α for all n ≥ n₀, and then ∞ if {a_n} does not tend to ∞. Similarly, we say that {a_n} tends to −∞ or diverges to −∞ if for every β ∈ ℝ, there is n₀ ∈ ℕ such that a_n < β for all n ≥ n₀ , and we write a_n → −∞.

Theorem 2.2.4 Every convergent sequence in ℝ is bounded. The converse is not true.

Proof. Let the sequence {a_n} converge to the limit ℓ. If ε = 1, there exists a positive integer N such that

The triangle inequality yields

it follows |a_n| ≤ 1 + |ℓ| for all n > N. Now, set

Thus |a_n| ≤ K for all n ∈ ℕ. In other words {a_n} is bounded by −K and K.  □

Is the converse of the above theorem true? No, as the following example shows. We will show later that a bounded monotone sequence converges.

Exmaple 2.2.5 {(−1)ⁿ)} is bounded by 1 but does not have a unique limit. Suppose that (−1)ⁿ ⟶ ℓ as n ⟶ ∞. Given ε = 1/2, then there is an N ∈ ℕ such that n ≥ N implies |(−1)ⁿ − ℓ| < ε For n odd this implies |1 + ℓ| = |−1 −ℓ| < 1/2 and for n even this implies |1 − ℓ| < 1/2. Hence

This implies 2 < 1, a contradiction. Therefore, the sequence (−1)ⁿ does not converge.  ■

Our recipe for finding the limit of a sequence {a_n} is therefore to show that {a_n} is monotone and bounded.

The Algebra of limits

Theorem 2.2.6. Suppose lim_{n ⟶ ∞} a_n = a and lim_{n ⟶ ∞} b_n = b. Then

1. the sequence {a_n ± b_n} are convergent, and lim_{n ⟶ ∞} a_n ± b_n = a ± b;

2. the sequence {a_nb_n} is convergent, and lim_{n ⟶ ∞} a_nb_n = ab;

3. if b ≠ 0 and {b_n} ≠ 0 for n ∈ ℕ, {a_n/ b_n} convergent, and lim_{n ⟶ ∞} a_n/b_n = a/b.

Proof. (a) To prove that a_n + b_n ⟶ a + b, we must show that |(a_n + b_n) − (a + b| can be made small by making n large. By the triangular inequality

This inequality suggests that, given ε > 0, since {a_n} converges to a we can find n ≥ n₁ such that |a_n − a| < ε/2 and similarly since {b_n} converges to b we can find n ≥ n₂ such that |b_n − b| < ε/2. Then for any n ≥ n0 = max{n₁, n₂}, both inequalities must hold, and so we have

b) To prove a_nb_n ⟶ ab, we attempt as above to relate the quantity |a_nb_n − ab| to the quantities |a_n − a| and |b_n − b|. To this scope we add and subtract the term a_nb

Since a_n ⟶ a, |a_n − a| < ε; and by the triangle inequality |a_n| < |a| + ε; Since b_n ⟶ b, |b_n − b| < ε; Thus

from the arbitrary of ε we have the claim.

c) We have

$$|\frac{a_n}{b_n}-\frac{a}{b}|=\frac{|a_{n}b-a{b}_n|}{|b_{n}b|}\le \frac{|a||b_n-b|+|b||a_n-b|}{|b_n|\cdot |b|}$$

since |a_n − a| < ε; and |b_n − b| < ε; and by the triangle inequality |b_n| < |b| + ε we have

$$|\frac{a_n}{b_n}-\frac{a}{b}|<\epsilon \frac{|a|+|b|}{|b{|}^2}$$

from the arbitrary of ε the claim follows.  □

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