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Bounded Sequence

Bounded Sequence

A sequence with terms that have an upper bound and a lower bound. For example, the harmonic sequence The harmonic sequence: 1, 1/2, 1/3, 1/4, … is bounded since no term is greater than 1 or less than 0.

Key Formula

A sequence {an} is bounded if there exist real numbers m and M such that m≤an≤M for all n.\text{A sequence } \{a_n\} \text{ is bounded if there exist real numbers } m \text{ and } M \text{ such that } m \leq a_n \leq M \text{ for all } n.
Where:
  • ana_n = The nth term of the sequence
  • mm = A lower bound — a number that is less than or equal to every term
  • MM = An upper bound — a number that is greater than or equal to every term
  • nn = The index (position) of each term, typically starting at 1

Worked Example

Problem: Determine whether the sequence a_n = (-1)^n · (3/n) is bounded.
Step 1: Write out several terms to see the pattern.
a1=−3,a2=32,a3=−1,a4=34,a5=−35,…a_1 = -3,\quad a_2 = \frac{3}{2},\quad a_3 = -1,\quad a_4 = \frac{3}{4},\quad a_5 = -\frac{3}{5}, \ldots
Step 2: Find the absolute value of each term. Since |(-1)^n| = 1, we have |a_n| = 3/n. The largest absolute value occurs at n = 1, giving |a_1| = 3.
∣an∣=3n≤3 for all n≥1|a_n| = \frac{3}{n} \leq 3 \text{ for all } n \geq 1
Step 3: This means every term satisfies -3 ≤ a_n ≤ 3. We can choose m = -3 and M = 3.
−3≤an≤3 for all n-3 \leq a_n \leq 3 \text{ for all } n
Step 4: Since both a lower bound and an upper bound exist, the sequence is bounded.
Answer: The sequence a_n = (-1)^n · (3/n) is bounded, with lower bound m = -3 and upper bound M = 3.

Another Example

Problem: Is the sequence a_n = 2n − 1 (i.e., 1, 3, 5, 7, …) bounded?
Step 1: Observe that the terms grow without limit: as n increases, 2n − 1 increases without end.
a1=1,a10=19,a100=199,a1000=1999,…a_1 = 1,\quad a_{10} = 19,\quad a_{100} = 199,\quad a_{1000} = 1999, \ldots
Step 2: A lower bound exists — every term is at least 1. However, no finite number M can satisfy a_n ≤ M for all n, because for any proposed M, choosing n large enough gives 2n − 1 > M.
For any M, choose n>M+12  ⟹  an=2n−1>M\text{For any } M, \text{ choose } n > \frac{M+1}{2} \implies a_n = 2n - 1 > M
Step 3: Since no upper bound exists, the sequence is not bounded (it is unbounded).
Answer: The sequence a_n = 2n − 1 is not bounded because it has no upper bound.

Frequently Asked Questions

Is every convergent sequence bounded?
Yes. If a sequence converges to a limit L, then eventually all terms are close to L, and the finitely many remaining terms are each finite. So you can always find an upper bound and a lower bound that contain every term. However, the reverse is not true — a bounded sequence does not have to converge (for example, (-1)^n is bounded but does not converge).
What is the difference between bounded above and bounded?
A sequence is bounded above if there exists some number M with a_n ≤ M for all n. It is bounded below if there exists some number m with a_n ≥ m for all n. A sequence is bounded only when it is both bounded above and bounded below.

Bounded Sequence vs. Convergent Sequence

A convergent sequence always approaches a specific limit, and every convergent sequence is necessarily bounded. A bounded sequence, however, does not have to converge. The classic example is an=(−1)na_n = (-1)^n, which is bounded (between −1-1 and 11) yet oscillates forever and never settles on one value. The Monotone Convergence Theorem bridges these ideas: a sequence that is both bounded and monotonic must converge.

Why It Matters

Boundedness is one of the first properties you check when analyzing a sequence. It is essential in calculus and analysis: the Bolzano–Weierstrass theorem guarantees that every bounded sequence has a convergent subsequence, and the Monotone Convergence Theorem states that a bounded, monotonic sequence must converge. These results are foundational tools for proving limits exist and for studying infinite series.

Common Mistakes

Mistake: Assuming that a bounded sequence must converge.
Correction: Bounded does not imply convergent. The sequence (-1)^n is bounded between -1 and 1 but does not converge. You need an additional condition, such as monotonicity, to guarantee convergence.
Mistake: Checking only an upper bound or only a lower bound and concluding the sequence is bounded.
Correction: A sequence is bounded only if both an upper bound and a lower bound exist. For instance, a_n = (-1)^n · n has no upper bound (it takes arbitrarily large positive values) and no lower bound (it takes arbitrarily large negative values), even though individual subsequences might appear bounded on one side.

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