Fundamental theorems on sequences

Theorem 2.2.1 (Uniqueness of the limit). If {an} is a convergent sequence with limit ℓ, there are no other limits.

Proof. Suppose there exist two limits ℓ1 and ℓ2 for the same sequence. By using triangular inequality:

|12| = |1an + an2| ≤ |1an| + |an2| < 2ε

Since ε is a real positive number (can be chosen to be arbitrarily small), this inequality can only hold if 1 = 2.  □

Theorem 2.2.2. (Squeeze Theorem). Let {an}, {bn} and {cn} be sequences such that

anbncn

for every positive integer n. If

limn ⟶ ∞ an = = limn ⟶ ∞ cn

then

limn ⟶ ∞ bn =

Proof. Let ε > 0. By the definition of limit, there exists a positive integer N1 such that if nN1:

ε < an <

An there exists a positive integer N2 such that if nN2:

ℓ − ε < cn < + ε

Thus is n ≥ max {N1,N2}, we have

ε < anbncn < + ε  □

Example 2.2.3. To illustrate the squeeze theorem, we find the limit of the following sequence (sin n)/n as n → ∞. Recall that the sine function has range [−1, +1] so

−1 ≤ sin n ≤ 1

for every value of n; when we divide these inequalities by n, we obtain

−1/n ≤ (sin n)/n ≤ 1/n

The flanking sequences are −1∕n and 1∕n, and both of them converge to 0 as n → ∞. Therefore, {an} converges to 0 too.  ■

Theorem 2.2.4. (Constancy of sign) If a sequence {an} has limit ≠ 0, its terms starting from an index ν on, have the same sign of ℓ.

Proof. Let ε= /2. By the definition of limit:

− ||/2 < an < + ||/2

n > ν, given a certain ν (whose value depends on the choice of ε). Thus if > 0 it follows (from the first inequality)

an > /2 > 0

or if ℓ < 0 it follows (from the second inequality)

an < ℓ/2 < 0. □

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