Properties of Sequences

12 min read

Textbook Reference

Two earlier skills come first: listing the terms of a sequence, and reasoning about its limit.

Key idea

Sometimes you can show a sequence converges without ever computing its limit.

Here is the picture. Suppose a sequence only ever climbs (each term is larger than the one before), and suppose it has a ceiling it never crosses. Then the terms are trapped: they keep rising but cannot escape past the ceiling. Geometrically, they are forced to crowd together against some level at or below the ceiling. They must settle. That level is the limit, and you know it exists even if you cannot name its value yet.

Two simple shape properties, “always heads one direction” (monotonic) and “stays inside a box” (bounded), combine into a strong guarantee: a sequence that is both must converge. The payoff is large, because many sequences, especially ones defined by a recursion, are easy to show monotonic and bounded but hard to take a direct limit of. This is the tool that handles them.

Prerequisite Check

• List terms of a sequence, including a recursively defined one (first lesson)
• Understand what $\lim_{n\to\infty} a_n = L$ means (second lesson)
• Compare two fractions to decide which is larger
• Take a derivative $f'(x)$ and read its sign to tell whether $f$ rises or falls
• Follow a short proof by mathematical induction

If these are solid, you are ready.

Quick Reference

Monotonic. A sequence $\{a_n\}$ is increasing if $a_1 < a_2 < a_3 < \cdots$ (every term larger than the last) and decreasing if $a_1 > a_2 > a_3 > \cdots$. It is monotonic if it is one or the other.

Bounded. A sequence is bounded above if some number $M$ satisfies $a_n \le M$ for all $n$, and bounded below if some number $m$ satisfies $m \le a_n$ for all $n$. It is bounded if it is both bounded above and below.

Monotonic Sequence Theorem. Every bounded, monotonic sequence converges. In particular, an increasing sequence that is bounded above converges, and a decreasing sequence that is bounded below converges.

Two ways to show monotonic.

1. Compare consecutive terms. Show $a_{n+1} > a_n$ (increasing) or $a_{n+1} < a_n$ (decreasing) directly, often by cross-multiplying.
2. Use the related function. If $f(x)$ with $f(n) = a_n$ has $f'(x) > 0$ on $[1, \infty)$, the sequence is increasing; if $f'(x) < 0$, it is decreasing.

Caution. Bounded alone does not give convergence ($\{(-1)^n\}$ is bounded but diverges), and monotonic alone does not either ($a_n = n$ is increasing but runs to infinity). You need both.

Key Concepts

1. Monotonic Sequences

A monotonic sequence never reverses direction. Once it starts climbing it keeps climbing; once it starts falling it keeps falling.

Example 1 (decreasing by inspection). Show that $a_n = \dfrac{3}{n+5}$ is decreasing.

Goal. Show each term is smaller than the one before, $a_{n+1} < a_n$.

Work. As $n$ grows, the denominator $n+5$ grows, so the fraction $\dfrac{3}{n+5}$ shrinks. Concretely, \[ a_n = \frac{3}{n+5} > \frac{3}{n+6} = \frac{3}{(n+1)+5} = a_{n+1} \] for all $n \ge 1$.

Answer: the sequence is decreasing.

Recap. When a positive numerator sits over a growing denominator, the terms shrink. Comparing $a_n$ with $a_{n+1}$ directly is the cleanest argument here.

Example 2 (the two standard methods on one sequence). Show that $a_n = \dfrac{n}{n^2+1}$ is decreasing.

Goal. Demonstrate both Stewart methods so you can choose whichever fits a given problem.

Method 1: compare consecutive terms. We must show $a_n > a_{n+1}$, that is, \[ \frac{n}{n^2+1} > \frac{n+1}{(n+1)^2+1}. \] Cross-multiplying (both denominators are positive) gives the equivalent inequality \[ n\big((n+1)^2+1\big) > (n+1)(n^2+1). \] Expanding both sides, \[ n^3 + 2n^2 + 2n > n^3 + n^2 + n + 1 \quad\Longleftrightarrow\quad n^2 + n > 1, \] which is true for every $n \ge 1$. So $a_n > a_{n+1}$ and the sequence is decreasing.

Method 2: use the related function. Let $f(x) = \dfrac{x}{x^2+1}$. By the quotient rule, \[ f'(x) = \frac{(x^2+1) - x(2x)}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2}, \] which is negative whenever $x^2 > 1$, that is for $x > 1$. So $f$ is decreasing on $(1, \infty)$, hence $f(n) > f(n+1)$, hence $\{a_n\}$ is decreasing.

Answer: the sequence is decreasing (both methods agree).

Recap. Method 1 is pure algebra; Method 2 trades the algebra for one derivative. Use Method 2 when the derivative is easy and Method 1 when cross-multiplying is cleaner.

(Common error: stopping after checking one or two terms. Seeing $a_1 > a_2 > a_3$ is suggestive but is not a proof. A sequence can decrease for a while and then turn around. You must show $a_{n+1} < a_n$ for all $n$, or show $f'(x) < 0$ on the whole interval.)

2. Bounded Sequences

A sequence is bounded when it stays inside a horizontal strip: it never rises above some ceiling $M$ and never drops below some floor $m$.

For instance, $a_n = n$ is bounded below (every term exceeds $0$) but not above (it grows forever). The sequence $a_n = \dfrac{n}{n+1}$ is bounded, because $0 < a_n < 1$ for all $n$: the floor is $0$ and the ceiling is $1$.

A bound does not have to be tight or “reached.” For $a_n = \dfrac{n}{n+1}$, the ceiling $M = 1$ works even though no term equals $1$. Any number that the terms never cross is a valid bound.

3. The Monotonic Sequence Theorem

Neither property alone forces convergence, but together they do.

Monotonic Sequence Theorem. Every bounded, monotonic sequence is convergent. In particular, a sequence that is increasing and bounded above converges, and a sequence that is decreasing and bounded below converges.

Why it is true, in words: if the terms always climb but can never pass a ceiling, they are squeezed into a shrinking room and must pile up against some level $L$. (The formal proof rests on the Completeness Axiom for the real numbers, the statement that the real line has no gaps, so a set with an upper bound has a least upper bound. That least upper bound is the limit.)

The strength of this theorem is that it certifies a limit exists before you know its value. That is exactly what you need for recursively defined sequences.

Example 3 (a recursive sequence shown convergent). Investigate the sequence defined by \[ a_1 = 2, \qquad a_{n+1} = \tfrac{1}{2}(a_n + 6). \]

Goal. Show the sequence is increasing and bounded above, conclude it converges by the theorem, then find the limit.

Work. The first terms are $2, 4, 5, 5.5, 5.75, 5.875, \dots$, which suggest the sequence is increasing and approaching $6$.

Step 1: increasing (by induction). The claim $a_{n+1} > a_n$ holds at $n = 1$ since $a_2 = 4 > 2 = a_1$. Assume $a_{k+1} > a_k$. Then $a_{k+1} + 6 > a_k + 6$, and dividing by $2$, \[ \tfrac{1}{2}(a_{k+1}+6) > \tfrac{1}{2}(a_k+6), \quad\text{that is}\quad a_{k+2} > a_{k+1}. \] By induction the sequence is increasing.

Step 2: bounded above by $6$ (by induction). The claim $a_n < 6$ holds at $n=1$ since $a_1 = 2 < 6$. Assume $a_k < 6$. Then $a_k + 6 < 12$, so \[ a_{k+1} = \tfrac{1}{2}(a_k+6) < \tfrac{1}{2}(12) = 6. \] By induction $a_n < 6$ for all $n$.

Step 3: the limit exists, so solve for it. The sequence is increasing and bounded above, so by the Monotonic Sequence Theorem the limit $L = \lim_{n\to\infty} a_n$ exists. Because $a_{n+1} \to L$ as well, take the limit of both sides of the recursion: \[ L = \tfrac{1}{2}(L + 6) \;\Longrightarrow\; 2L = L + 6 \;\Longrightarrow\; L = 6. \]

Answer: the sequence converges, and $\displaystyle\lim_{n\to\infty} a_n = 6$.

Recap. Notice the order. You cannot solve $L = \tfrac{1}{2}(L+6)$ for the limit until you first know the limit exists, and the theorem is what supplies that. Monotonic-and-bounded earns you the right to take the limit of the recursion.

(Common error: jumping straight to $L = \tfrac{1}{2}(L+6)$ without first proving the sequence converges. If the sequence diverged, that equation would be meaningless. The two induction steps are not optional; they justify the final algebra.)

Inline Self-Check

Question. A sequence is increasing and every term lies between $5$ and $8$. Must it converge? What can you say about its limit $L$?

Common Errors Summary

Common Misconceptions

Leveled Practice

Attempt each problem before opening the answer.

Level 1: Direct Application

Problem 1. Show that $a_n = \dfrac{1}{2n+3}$ is decreasing.

Problem 2. Is $a_n = \dfrac{1-n}{2+n}$ increasing, decreasing, or neither? Is it bounded?

Problem 3. Give a ceiling and a floor for $a_n = \dfrac{n}{n+4}$.

Level 2: Both Methods

Problem 4. Show that $a_n = \dfrac{2n+3}{3n+4}$ is decreasing using the related-function method.

Problem 5. Decide whether $a_n = n + \dfrac{1}{n}$ is monotonic, and whether it is bounded.

Level 3: Recursive Sequences

Problem 6. The sequence is defined by $a_1 = \sqrt{2}$ and $a_{n+1} = \sqrt{2 + a_n}$. It can be shown to be increasing and bounded above by $3$. Granting that, find $\displaystyle\lim_{n\to\infty} a_n$.

Problem 7. Show that the sequence $a_1 = 1$, $a_{n+1} = 3 - \dfrac{1}{a_n}$ is increasing and bounded above by $3$, then find its limit.

Mastery Checklist

You have mastered this skill when you can do all of the following without notes:

• Define increasing, decreasing, and monotonic
• Define bounded above, bounded below, and bounded
• Show a sequence is monotonic by comparing consecutive terms
• Show a sequence is monotonic using the sign of the related function’s derivative
• State the Monotonic Sequence Theorem and explain why both hypotheses are needed
• Use the theorem to prove a recursive sequence converges before finding the limit
• Find the limit of a convergent recursive sequence by taking the limit of the recursion

Mental Model

Think of the terms as a ball rolling along a ramp inside a room with a ceiling.

Monotonic means the ramp only ever tilts one way: the ball only rolls up, or only rolls down, never reversing. Bounded means the room has a ceiling and a floor the ball never passes. Put the two together and the ball is rolling steadily upward but cannot break through the ceiling. It has nowhere to go but to ease toward some level and stop. That stopping level is the limit.

Drop either condition and the guarantee fails. Remove the ceiling (not bounded) and the ball rolls up forever. Remove the steady tilt (not monotonic) and the ball can rattle around between the floor and ceiling without ever settling. Both conditions together are what pin the ball down.

Connections

Built From

• Sequences and notation: you list terms, including recursive ones, to see the behavior.
• Limits of sequences: the theorem concludes that a limit exists, so it only makes sense once you know what a limit is.

Leads To

• Infinite series and the divergence test: a series is the sequence of its partial sums, and for a series of positive terms that partial-sum sequence is increasing. Whether it is bounded is exactly the question of whether the series converges. The monotonic-and-bounded idea is the engine behind the integral test and the comparison tests later in the chapter.

Why It Matters

The ability to prove a limit exists without computing it is what makes iterative methods trustworthy. A successive-approximation scheme, such as Newton’s method or a fixed-point iteration, produces a recursive sequence; showing it is monotonic and bounded proves it converges to an answer even before that answer is known in closed form. In numerical analysis this is the standard route to a convergence guarantee, and the final “take the limit of the recursion” step is exactly how the fixed point is identified.

Audience Notes

For students who find math intimidating: you do not have to find the limit to know one exists. If a sequence only ever climbs and has a ceiling, it must settle. That single idea answers many problems that look hard.

For students who want depth: the theorem rests on the Completeness Axiom, the property that distinguishes the real numbers from the rationals (the rationals have gaps; the reals do not). Tracing why completeness is exactly what is needed here is a first real taste of analysis.

For students aimed at a career in computing or engineering: a recursive sequence is a fixed-point iteration. Proving it is monotonic and bounded is how you certify that an iterative solver converges, and taking the limit of the recursion is how you find the fixed point it converges to.

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