Bounded Sequences: Definition, Sup, Inf and Examples

What Are Bounded Sequences?

Bounded sequences are sequences whose terms all stay inside one fixed interval, so no term ever climbs above some ceiling $M$ or falls below some floor $m$. A sequence is an ordered list $a_1, a_2, a_3, \ldots$, one term for each positive integer $n$. Being bounded says nothing yet about where the terms go; it only promises that they never run off to plus or minus infinity.

The formal definitions come in three layers. A sequence ${a_n}$ is bounded above if there is a real number $M$ with

$$a_n \leq M \quad \text{for every } n \geq 1.$$

It is bounded below if there is a real number $m$ with

$$a_n \geq m \quad \text{for every } n \geq 1.$$

A bounded sequence is one that is bounded above and bounded below at the same time:

$$m \leq a_n \leq M \quad \text{for every } n \geq 1.$$

Here $M$ is an upper bound, $m$ is a lower bound, and both are fixed numbers that do not depend on $n$. That last condition matters. A "bound" such as $a_n \leq n$ changes with $n$, so it proves nothing.

Why Is The Absolute-Value Form The Same As Being Bounded?

Ask "What is meant by bounded sequence?" and most textbooks answer with a single inequality with the absolute value:

$$|a_n| \leq K \quad \text{for every } n \geq 1, \text{ where } K > 0.$$

The two versions say the same thing. If $|a_n| \leq K$, then $-K \leq a_n \leq K$, so $m = -K$ and $M = K$ work. Going the other way, if $m \leq a_n \leq M$, take $K = \max(|m|, |M|)$, and every term satisfies $|a_n| \leq K$. The absolute-value form is quicker in proofs; the two-sided form tells you more, because the floor and the ceiling can be very different numbers.

How Do Bounded Above, Bounded Below, And Bounded Differ?

A sequence can have a ceiling with no floor, a floor with no ceiling, both, or neither. "What does it mean for a sequence to be bounded above and below, and what are some examples?" is best answered with a side-by-side table, because the same handful of sequences turns up in every course.

Read the table column by column and three patterns jump out. Every sequence that converges is bounded. Several bounded sequences still fail to converge. And the supremum or infimum is often a value the sequence approaches but never actually hits.

What Are The Supremum And Infimum Of A Sequence?

The supremum of a sequence is its least upper bound: the smallest number that is still at or above every term. The infimum is its greatest lower bound: the largest number that is still at or below every term. Students ask "What is bounded, upper bound, and least upper bound?" because a bounded sequence has infinitely many upper bounds (if 2 works, so do 3, 10 and 1000), and the supremum picks out the single tightest one.

In symbols, $s = \sup{a_n}$ means two things at once:

$$a_n \leq s \text{ for every } n, \qquad \text{and for every } \varepsilon > 0 \text{ some term satisfies } a_n > s - \varepsilon.$$

The first line says $s$ is a ceiling. The second says nothing lower works, because any lower ceiling $s - \varepsilon$ gets pierced by some term. The infimum mirrors this with the inequalities reversed.

Is The Supremum The Same As The Largest Term?

Only sometimes. For $\frac{1}{n}$ the supremum $1$ is the first term, so it is also the maximum. For $1 - \frac{1}{n}$ the supremum is again $1$, but no term ever equals $1$: the terms $0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \ldots$ creep toward it forever.

A maximum must be a term; a supremum need not be. This distinction is genuinely slippery: "least upper bound" sounds as if it should mean the biggest term, and the wording hides the difference.

Why does a bounded sequence always have a supremum at all? That is the completeness property of the real numbers: every non-empty set of reals with an upper bound has a least upper bound. The rational numbers fail this test (the decimals $1, 1.4, 1.41, 1.414, \ldots$ have no rational supremum, since $\sqrt{2}$ is irrational), which is the deepest reason calculus is done over the reals.

How Do You Prove That A Sequence Is Bounded?

To prove a sequence is bounded, you exhibit two fixed numbers $m$ and $M$ and show every term lies between them. The question "How to know if a sequence is bounded?" has no single recipe, but five methods cover almost every school and first-year problem:

• Rewrite the formula. Split a fraction so its behaviour is visible: $\frac{2n+3}{n+1} = 2 + \frac{1}{n+1}$, and since $0 < \frac{1}{n+1} \leq \frac{1}{2}$, every term lies between $2$ and $2.5$.

• Use standard inequalities. Facts such as $-1 \leq \sin n \leq 1$ and the triangle inequality $|x + y| \leq |x| + |y|$ give bounds in one line: $\left|\frac{\sin n}{n}\right| \leq \frac{1}{n} \leq 1$.

• Monotonic plus one end. If the sequence is decreasing, the first term is an upper bound for free, so you only need a lower bound (and vice versa for increasing sequences).

• Induction for recursive sequences. When $a_{n+1}$ is defined from $a_n$, prove "if $a_n < M$ then $a_{n+1} < M$" and check the first term. This is mathematical induction doing the work, and it is the only practical route for formulas without a closed form.

• Show it converges. Every convergent sequence is bounded (proved in the next section), so a known limit settles boundedness at once.

Notice what is missing from the list: checking the first ten terms. A finite table of values can never prove a bound, because the trouble may start at term one thousand.

Is Every Convergent Sequence Bounded?

Yes. If $a_n \to L$, the sequence is bounded. This half of the classic pair "Is every bounded sequence convergent? Is every convergent sequence bounded?" has a short proof worth knowing, because it shows how a limit turns into a bound.

Take $\varepsilon = 1$ in the definition of the limit of a sequence. Then there is an index $N$ with

$$|a_n - L| < 1 \quad \text{for all } n > N,$$

which rearranges to

$$|a_n| < |L| + 1 \quad \text{for all } n > N.$$

That handles the infinite tail. The head $a_1, a_2, \ldots, a_N$ is a finite list, so it has a largest absolute value. Set

$$K = \max\big(|a_1|, |a_2|, \ldots, |a_N|, ; |L| + 1\big),$$

and every term satisfies $|a_n| \leq K$. The choice $\varepsilon = 1$ is arbitrary (any fixed positive number works), a point that confuses many first readers of this proof; one is simply the tidiest choice.

The contrapositive is the useful daily tool: an unbounded sequence cannot converge. The moment you show $n^2$ or the Fibonacci numbers $1, 1, 2, 3, 5, 8, \ldots$ exceed every proposed ceiling, divergence follows.

Is Every Bounded Sequence Convergent?

No. Bounded is necessary for convergence but not sufficient. The standard counterexample is $a_n = (-1)^n$, which lists $-1, 1, -1, 1, \ldots$: it stays inside $[-1, 1]$ forever but never settles near a single number, so it has no limit.

A bounded sequence can fail to converge in a few ways:

• It oscillates between values. $(-1)^n$ jumps between two points.

• It oscillates between clusters. $(-1)^n + \frac{1}{n}$ has its even terms drifting toward $1$ and its odd terms drifting toward $-1$.

• It wanders. $\sin n$ visits every part of $[-1, 1]$ without settling anywhere.

What extra condition would rescue convergence? Two classical answers exist, and they are the next two sections.

Why Does An Increasing Bounded Sequence Converge?

The first rescue is direction. The Monotone Convergence Theorem says that a sequence which is both monotonic and bounded must converge: an increasing sequence that is bounded above converges to its supremum, and a decreasing sequence that is bounded below converges to its infimum. The theorem holds even if the sequence only becomes monotonic after some index $n_0$.

The question "Why does the limit of an increasing bounded sequence converge?" has a satisfying answer once the supremum is in hand. Let $s = \sup{a_n}$ for an increasing, bounded-above sequence.

For any $\varepsilon > 0$, some term $a_N$ exceeds $s - \varepsilon$ (that is what "least" upper bound means). Because the sequence increases, every later term is at least $a_N$, and none passes $s$. So from $N$ onward every term is squeezed into the gap between $s - \varepsilon$ and $s$, which is exactly the statement $a_n \to s$.

Testing whether a sequence is increasing (difference, ratio or derivative test) is its own skill, covered with worked examples in monotonic sequences. Here the point is narrower: boundedness supplies the ceiling, monotonicity supplies the direction, and together they force a limit even when you cannot compute it.

What Does The Bolzano-Weierstrass Theorem Say About Bounded Sequences?

The second rescue gives up on the whole sequence and settles for part of it. The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence, so the exam-style blank "Every bounded sequence has ..." is filled by "a convergent subsequence".

A subsequence keeps infinitely many terms in their original order and throws the rest away. For $(-1)^n$, keeping only the even-indexed terms gives $1, 1, 1, \ldots$, which converges to $1$. For $(-1)^n + \frac{1}{n}$, the even terms converge to $1$ and the odd terms to $-1$. The theorem guarantees that this always works, even for something as erratic as $\sin n$, where nobody writes the subsequence down explicitly.

The usual proof is a halving argument: trap infinitely many terms in $[m, M]$, cut the interval in half, keep a half that still holds infinitely many terms, and repeat. The nested intervals shrink to a single point, and picking one term from each interval builds a subsequence that converges to it. Full rigour belongs to a real-analysis course; at school level, the statement and the $(-1)^n$ picture are what you need.

Can A Sequence Be Unbounded Without Going To Infinity?

Yes, and this gap trips up many students. "Unbounded" and "diverges to infinity" are different claims. A sequence diverges to $+\infty$ only if its terms eventually exceed every number and stay above it. The sequence

$$a_n = (-1)^n, n: \quad -1, ; 2, ; -3, ; 4, ; -5, ; 6, ; \ldots$$

has no upper bound and no lower bound, yet it does not tend to $+\infty$ or to $-\infty$, because it keeps swinging back across zero. It simply diverges.

So the honest classification of a real sequence has more boxes than "converges or blows up":

• converges (always bounded);

• bounded but divergent, like $(-1)^n$;

• diverges to $+\infty$ or $-\infty$ (unbounded on one side, always);

• unbounded with no infinite limit, like $(-1)^n, n$.

Examples Of Bounded Sequences

Example 1

Show that $a_n = \frac{3}{n+2}$ is bounded, and find its supremum and infimum.

Every term is positive, so $0$ is a lower bound:

$$a_n = \frac{3}{n+2} > 0.$$

The denominator grows with $n$, so the terms decrease, and the largest is the first:

$$a_1 = \frac{3}{3} = 1.$$

So $0 < a_n \leq 1$ for every $n$, and the limit is

$$\lim_{n \to \infty} \frac{3}{n+2} = 0.$$

Final answer: bounded; $\sup = 1$ (reached at $n = 1$), $\inf = 0$ (never reached).

Example 2

Find the supremum and infimum of $a_n = (-1)^n + \frac{1}{n}$.

A tempting shortcut: the supremum of $(-1)^n$ is $1$ and the supremum of $\frac{1}{n}$ is $1$, so add them:

$$\sup a_n = 1 + 1 = 2 \quad (\text{wrong}).$$

Check it against actual terms. For the sum to reach $2$, both parts would need to hit their maximum at the same $n$, but $(-1)^n = 1$ needs $n$ even, while $\frac{1}{n} = 1$ needs $n = 1$, which is odd. The first few terms show no term near $2$:

$$a_1 = 0, \quad a_2 = \frac{3}{2}, \quad a_3 = -\frac{2}{3}, \quad a_4 = \frac{5}{4}, \quad a_5 = -\frac{4}{5}.$$

Now split by parity. For even $n$, $a_n = 1 + \frac{1}{n}$, which is largest at $n = 2$:

$$a_2 = 1 + \frac{1}{2} = \frac{3}{2}.$$

For odd $n$, $a_n = -1 + \frac{1}{n}$, which decreases toward $-1$ and never reaches it:

$$\lim_{n \to \infty} \left(-1 + \frac{1}{n}\right) = -1.$$

Adding suprema only gives an upper bound ($2$ is a ceiling), not the least one.

Final answer: $\sup = \frac{3}{2}$ (reached at $n = 2$), $\inf = -1$ (never reached). The sequence is bounded but does not converge.

Example 3

Prove that $a_n = \frac{2n+3}{n+1}$ is bounded and decide whether it converges.

Rewrite the fraction:

$$a_n = \frac{2n+3}{n+1} = 2 + \frac{1}{n+1}.$$

Since $0 < \frac{1}{n+1} \leq \frac{1}{2}$ for $n \geq 1$:

$$2 < a_n \leq \frac{5}{2}.$$

The difference of consecutive terms is negative:

$$a_{n+1} - a_n = -\frac{1}{(n+1)(n+2)} < 0,$$

so the sequence is decreasing and bounded below, and the Monotone Convergence Theorem gives a limit:

$$\lim_{n \to \infty} \left(2 + \frac{1}{n+1}\right) = 2.$$

Final answer: $2 < a_n \leq \frac{5}{2}$; $\sup = \frac{5}{2}$, $\inf = 2$ (never reached); converges to $2$.

Example 4

Show that $a_n = \frac{n}{2^n}$ is bounded and find its supremum.

Compare consecutive terms with a ratio:

$$\frac{a_{n+1}}{a_n} = \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} = \frac{n+1}{2n}.$$

For $n \geq 1$ this ratio is at most $1$ (it equals $1$ only at $n = 1$), so the terms never increase after the first:

$$a_1 = \frac{1}{2}, \quad a_2 = \frac{2}{4} = \frac{1}{2}, \quad a_3 = \frac{3}{8}.$$

All terms are positive, so $0$ is a lower bound, and

$$\lim_{n \to \infty} \frac{n}{2^n} = 0.$$

Final answer: $0 < a_n \leq \frac{1}{2}$; $\sup = \frac{1}{2}$ (reached twice, at $n = 1$ and $n = 2$), $\inf = 0$.

Example 5

Let $a_1 = 1$ and $a_{n+1} = \sqrt{2 + a_n}$. Show the sequence is bounded above by $2$, then find its limit.

Bound by induction. The base case holds since $a_1 = 1 < 2$. If $a_n < 2$, then

$$a_{n+1} = \sqrt{2 + a_n} < \sqrt{2 + 2} = 2.$$

So $1 \leq a_n < 2$ for every $n$. The sequence increases, because

$$a_{n+1} > a_n \iff 2 + a_n > a_n^2 \iff (a_n - 2)(a_n + 1) < 0,$$

which is true whenever $-1 < a_n < 2$. Increasing and bounded above means it converges to some $L$. Taking limits on both sides of the recursion:

$$L = \sqrt{2 + L}$$

$$L^2 - L - 2 = 0$$

$$(L - 2)(L + 1) = 0.$$

Since every term is positive, $L = 2$. The first terms $1, \sqrt{3} \approx 1.7321, 1.9319, 1.9829$ confirm the climb.

Final answer: bounded, $1 \leq a_n < 2$; the limit is $2$.

Example 6

A patient takes 100 mg of a drug every day, and half of the drug in the body is cleared between doses. With $a_1 = 100$ and $a_{n+1} = \frac{a_n}{2} + 100$ (in mg), show the amount never exceeds 200 mg and find the long-run level.

Bound by induction. The base case is $a_1 = 100 < 200$. If $a_n < 200$, then

$$a_{n+1} = \frac{a_n}{2} + 100 < \frac{200}{2} + 100 = 200.$$

The sequence increases, because

$$a_{n+1} - a_n = 100 - \frac{a_n}{2} > 0 \quad \text{whenever } a_n < 200.$$

A closed form confirms both facts:

$$a_n = 200 - 100\left(\frac{1}{2}\right)^{n-1},$$

giving $100, 150, 175, 187.5, 193.75, \ldots$, and

$$\lim_{n \to \infty} \left[200 - 100\left(\frac{1}{2}\right)^{n-1}\right] = 200.$$

Final answer: the amount is bounded, $100 \leq a_n < 200$ mg; it rises toward a long-run level of 200 mg, which is its supremum and is never reached. (This is a simplified model for illustration, not dosing advice.)

Why Do Mathematicians Care Whether A Sequence Is Bounded?

The reason is simple: "a bound is something you can prove before you know the limit." Many sequences, especially recursive ones like Example 5, have no formula you can take a limit of directly. A bound, plus a direction, proves that a limit exists, and only then does it make sense to solve $L = \sqrt{2 + L}$ for its value.

That idea sits under several larger structures:

• Existence without a formula. The Monotone Convergence Theorem and the Bolzano-Weierstrass theorem both start from a bound. They let mathematicians prove that solutions, maxima and limits exist before anyone can compute them.

• Series. An infinite series of positive terms converges exactly when its sequence of partial sums is bounded above. That single fact drives the comparison-style tests in convergence and divergence of series.

• Stability. In engineering, a system is called stable in the bounded-input, bounded-output sense when every bounded input signal produces a bounded output signal. An audio amplifier or a cruise-control loop whose output sequence grows without bound is a system that has failed.

The Mathematicians Behind Bounded Sequences

Bernard Bolzano (1781 to 1848, Prague, Bohemia) published a paper in 1817 that proved the intermediate value theorem by purely arithmetic arguments instead of geometric intuition. Inside that proof he established, as a lemma, the result now known as the Bolzano-Weierstrass theorem.

Karl Weierstrass (1815 to 1897, Westphalia, Germany) taught analysis at Berlin with a standard of rigour that still shapes how the subject is taught. Roughly fifty years after Bolzano, he identified the bounded-sequence lemma as important in its own right and proved it again, which is why both names are attached to it.

Where Are Bounded Sequences Used In Real Life?

Bounded sequences show up wherever a repeated process has to stay inside safe limits:

• Repeated dosing. Example 6 is the simplest version of how pharmacologists reason about a drug building up toward a steady level instead of rising indefinitely.

• Numerical methods. Iterative algorithms such as Newton's method produce a sequence of approximations; showing the iterates stay in a bounded interval is a standard first step in proving they converge.

• Population models. In the logistic map $x_{n+1} = r,x_n(1 - x_n)$ with $0 \leq r \leq 4$, every term stays between $0$ and $1$ when the starting value does, even in its chaotic range where it never converges.

Where Do Students Trip Up On Bounded Sequences?

Mistake 1: Treating "Bounded" As "Convergent"

Where it slips in: True or false items such as "All bounded sequences are convergent", and multiple-choice options that mix the two ideas.

Don't do this: Conclude that $(-1)^n$ or $\sin n$ has a limit because the terms stay between $-1$ and $1$.

The correct way: Treat the bound as half the evidence. Add monotonicity (Monotone Convergence Theorem) or find the limit directly; otherwise the most a bound guarantees is a convergent subsequence.

The memorizer is especially exposed here. A student who has memorised "bounded and monotonic implies convergent" often drops the second word under time pressure, and the theorem quietly becomes false.

Mistake 2: Calling The Supremum The Largest Term

Where it slips in: Finding \sup\left{1 - \frac{1}{n}\right} or any increasing sequence with a limit.

Don't do this: Write "the supremum is the biggest term, which is $a_{1000}$" or claim no supremum exists because no term is largest.

The correct way: Check both conditions: $s$ must be a ceiling, and nothing smaller may be a ceiling. For $1 - \frac{1}{n}$ the supremum is $1$ even though no term equals $1$.

Mistake 3: Judging A Bound From The First Few Terms

Where it slips in: Problems where the early terms are small and the growth starts late, such as $a_n = \frac{n^2}{n + 100}$.

Don't do this: List $a_1, \ldots, a_{10}$, see values from about $0.0099$ to $0.9091$, and declare "bounded by $1$".

The correct way: Prove the bound for every $n$, or disprove it. Here $a_{100} = \frac{10000}{200} = 50$, and

$$\lim_{n \to \infty} \frac{n^2}{n + 100} = \infty,$$

so the sequence is unbounded.

The rusher falls into this one. The first instinct on $\frac{n^2}{n+100}$ is to plug in small values of $n$ and trust the pattern, and the habit that fixes it is dividing top and bottom by $n$ before computing anything: $\frac{n}{1 + 100/n}$ makes the growth obvious at a glance.

Practice Problems On Bounded Sequences

1. Show that $a_n = \frac{5}{n+4}$ is bounded, and find its supremum and infimum.

2. Is $a_n = (-1)^n \frac{n}{n+1}$ bounded? Find its supremum and infimum, and decide whether it converges.

3. Show that $a_n = \frac{3n-1}{n+2}$ is increasing and bounded, and find its infimum, supremum and limit.

4. Let $a_1 = 1$ and $a_{n+1} = \sqrt{6 + a_n}$. Find an upper bound, show the sequence increases, and find its limit.

5. Find the infimum and supremum of $a_n = n \sin\left(\frac{1}{n}\right)$.

Answer to Question 1: $0 < a_n \leq 1$; $\sup = 1$ (at $n = 1$), $\inf = 0$ (never reached).

Answer to Question 2: Yes, $|a_n| < 1$. $\sup = 1$ and $\inf = -1$, neither reached (the terms are $-\frac{1}{2}, \frac{2}{3}, -\frac{3}{4}, \frac{4}{5}, \ldots$). It does not converge, since even terms approach $1$ and odd terms approach $-1$.

Answer to Question 3: $a_n = 3 - \frac{7}{n+2}$, and $a_{n+1} - a_n = \frac{7}{(n+2)(n+3)} > 0$. $\inf = a_1 = \frac{2}{3}$, $\sup = 3$ (never reached), limit $3$.

Answer to Question 4: By induction $a_n < 3$, since $a_n < 3$ gives $a_{n+1} < \sqrt{9} = 3$. It increases while $-2 < a_n < 3$ because $a_n^2 - a_n - 6 = (a_n - 3)(a_n + 2) < 0$. The limit solves $L = \sqrt{6 + L}$, so $L = 3$.

Answer to Question 5: The sequence increases, so $\inf = a_1 = \sin 1 \approx 0.8415$, and $\sup = 1$ (never reached), because $\lim_{n \to \infty} n \sin\left(\frac{1}{n}\right) = 1$.

What To Remember About Bounded Sequences

• Bounded sequences satisfy $m \leq a_n \leq M$ for fixed numbers $m$ and $M$, which is the same as $|a_n| \leq K$.

• The supremum is the least upper bound and the infimum is the greatest lower bound, and neither has to be a term.

• Every convergent sequence is bounded, but bounded sequences such as $(-1)^n$ can still diverge.

• Bounded plus monotonic forces convergence, and bounded alone still guarantees a convergent subsequence.

• A bound must be proved for every $n$; a table of early terms never proves one.

Your Next Move

Work through Practice Problems 1 to 5 in order; if Problem 4 trips you up, return to Example 5 and copy its induction structure line by line, and if Problem 2 does, reread the parity split in Example 2. Want a live Bhanzu trainer to walk through more bounded-sequence proofs? Book a free demo class.

Read More

• Infinite series and how partial sums behave

• Geometric series and their sums

• Why the harmonic series diverges

• Limits at infinity for functions

• Standard limits worth memorising