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Monotone and Bounded Sequences · Tutorial 528 of 998

What Numerical Terms Can And Cannot Show

Use numerical evidence to form useful conjectures about a sequence, while distinguishing what the calculations suggest from what requires proof.

Intermediate 9 min read

What You'll Learn

  • Read exact and decimal sequence terms without confusing rounded values with the terms themselves.
  • Use a finite sample to form conjectures about monotonicity and bounds.
  • Estimate a possible limit from the behaviour of computed terms.
  • Explain why checking finitely many terms cannot establish a property for every term.
  • Compare sequences with identical initial terms but different long-term behaviour.
  • Recognize when the Monotone Convergence Theorem can be used after its hypotheses are proved.

Numerical Evidence and Mathematical Proof

A list of computed terms is often the first clue to how a sequence behaves. The terms may appear to increase, remain within a fixed range, or approach a particular number. Those observations can help us form a conjecture and decide what to investigate next.

But a list is finite, while statements about a sequence usually concern every index or all sufficiently large indices. Even an accurate computation of many terms cannot, by itself, prove that a sequence is monotone, bounded, or convergent. The useful habit is to treat calculations as evidence for a conjecture—not as a substitute for a proof.

Key distinction: Computed terms show what happened at the indices we checked. A proof is needed to establish what happens at every relevant index, including those we have not checked.

What to Look For in a List of Terms

For a sequence \((a_n)\), the terms \(a_1,a_2,\ldots,a_N\) can suggest several different features:

  • Monotonicity: Do the displayed terms appear to rise or fall as \(n\) increases?
  • Bounds: Do the displayed terms appear to stay above or below a particular number?
  • A possible limit: Do the terms appear to settle near some value?

These are conjectures about the entire sequence, not conclusions established by the sample. A sequence is increasing if \(a_n\leq a_{n+1}\) for every \(n\in\mathbb N\); it is decreasing if \(a_n\geq a_{n+1}\) for every \(n\in\mathbb N\). A number \(M\) is an upper bound if \(a_n\leq M\) for every \(n\), and \(m\) is a lower bound if \(m\leq a_n\) for every \(n\). Convergence to \(L\) means that, for every \(\varepsilon>0\), there is an index \(N\) such that \(n\geq N\) implies \(|a_n-L|<\varepsilon\).

A sample can suggest candidates for these properties. It cannot check infinitely many inequalities or verify the definition of convergence for every required tolerance.

Worked Example: Terms Suggest an Increasing Sequence Below One

Worked Example: Terms Suggest an Increasing Sequence Below One

Consider \(a_n=\frac{n}{n+1}\). Compute several terms exactly, then give their decimal approximations:

$$ \begin{aligned} a_1&=\frac{1}{1+1}=\frac12=0.5,\\ a_2&=\frac{2}{2+1}=\frac23\approx0.6667,\\ a_3&=\frac{3}{3+1}=\frac34=0.75,\\ a_4&=\frac{4}{4+1}=\frac45=0.8,\\ a_5&=\frac{5}{5+1}=\frac56\approx0.8333,\\ a_{10}&=\frac{10}{10+1}=\frac{10}{11}\approx0.9091. \end{aligned} $$

In this sample, each displayed term is larger than the one before it, and every displayed term is less than \(1\). The terms also appear to get closer to \(1\). A reasonable conjecture is that the sequence increases, is bounded above by \(1\), and has possible limit \(1\).

Check: Substituting \(n=10\) into the defining formula gives \(a_{10}=10/11\), whose decimal approximation is \(0.9091\) to four decimal places. The calculation verifies that sampled term; it does not establish the conjectured properties for every \(n\).

Worked Example: A Sample Can Suggest a Limit and Bounds

Worked Example: A Sample Can Suggest a Limit and Bounds

Consider \(b_n=1+\frac1n\). The first five terms are:

$$ \begin{aligned} b_1&=1+\frac11=2,\\ b_2&=1+\frac12=\frac32=1.5,\\ b_3&=1+\frac13=\frac43\approx1.3333,\\ b_4&=1+\frac14=\frac54=1.25,\\ b_5&=1+\frac15=\frac65=1.2. \end{aligned} $$

These values appear to decrease, stay above \(1\), and approach \(1\). The sample therefore suggests \(1\) as a lower bound and possible limit. The largest displayed term is \(2\), so the sample also suggests \(2\) as an upper bound. These observations are not proofs that either bound holds for every term, or that the sequence decreases or converges.

Check: For the fifth term, substituting \(n=5\) gives \(b_5=1+1/5=6/5=1.2\). The exact fraction confirms the displayed decimal. As before, the check concerns a computed term, not the whole sequence.

Why a Finite List Cannot Prove a General Claim

Suppose the first \(N\) terms appear to have a particular pattern. Any claim such as “the sequence is increasing” or “the sequence is bounded above by \(M\)” concerns all indices, not just \(1,\ldots,N\). Terms after the sample might continue the pattern, or they might behave differently.

The same limitation applies to a proposed limit. The definition of convergence requires that, for every positive tolerance, all sufficiently late terms lie within that tolerance of the proposed limit. A finite list can show that certain terms lie close to a number. It cannot ensure that every later term will also lie close to it.

The point is stronger than saying that a short sample might be misleading. Even if we know an arbitrarily long finite initial segment exactly, there can be different sequences with that same segment and different behaviour afterward.

Worked Example: Identical Initial Terms, Different Behaviour

Worked Example: Identical Initial Terms, Different Behaviour

Let \(c_n=n/(n+1)\) for \(1\leq n\leq5\), so the first five terms are \(1/2, 2/3, 3/4, 4/5, 5/6\). Now define the rest of the sequence by

$$ c_n= \begin{cases} 0, & n\geq6\text{ and }n\text{ is even},\\ 2, & n\geq7\text{ and }n\text{ is odd}. \end{cases} $$

The first seven terms are therefore

$$ c_1=\frac12,\quad c_2=\frac23,\quad c_3=\frac34,\quad c_4=\frac45,\quad c_5=\frac56,\quad c_6=0,\quad c_7=2. $$

The first five terms match the earlier sample exactly. They rise toward \(1\); however, \(c_6=0\) is less than \(c_5=5/6\), and \(c_7=2\) is greater than \(c_6=0\). So the apparent increasing pattern does not continue. Moreover, from index \(6\) onward the terms alternate between \(0\) and \(2\), rather than settling near one number.

Check: At \(n=6\), the even-index rule gives \(c_6=0\); at \(n=7\), the odd-index rule gives \(c_7=2\). These substitutions verify the first two terms after the shared prefix and show directly that the sample pattern is not enough to describe what follows.

For comparison, a different sequence could keep the same first five terms and set every term from \(n=6\) onward equal to \(1\). That sequence would converge to \(1\), since all terms from index \(6\) onward would equal \(1\). The two sequences have identical first five terms, but their later behaviour differs. No calculation restricted to those five terms can distinguish between them.

A General Reason the Limitation Is Unavoidable

The preceding example expresses a general fact. Fix any number \(N\) of initial terms and any proposed value \(L\). After those first \(N\) terms, one can define a sequence to be constantly \(L\); that continuation converges to \(L\). One can instead define the later terms to alternate between \(L-1\) and \(L+1\); this continuation does not converge.

To see why the alternating continuation cannot converge, suppose it converged to some number \(A\). By the definition of convergence, eventually every term would be within \(1/2\) of \(A\). In particular, an even-indexed tail term \(L-1\) and an odd-indexed tail term \(L+1\) would both be within \(1/2\) of \(A\). The triangle inequality would then give

$$ 2=|(L+1)-(L-1)| \leq |(L+1)-A|+|A-(L-1)| <\frac12+\frac12=1, $$

which is impossible. Thus the two continuations agree on all the inspected terms but can differ in whether they converge. More computation may make a conjecture seem more plausible, but no finite prefix alone can prove convergence.

Consequence: Numerical evidence can guide a proof, but convergence is a statement about all sufficiently late terms. It cannot be established merely by extending a finite table.

From Conjecture to Proof

After computing terms, record what you suspect and what still needs justification. For example, a useful working summary might be:

1
Describe the observation.
The sampled terms appear to increase and remain below \(1\).
2
State the conjecture precisely.
Propose that the sequence is increasing and bounded above by \(1\).
3
Separate evidence from proof.
Check what would have to be established for every index, rather than only for the displayed terms.
4
Use an applicable theorem only after its hypotheses are proved.
For example, the Monotone Convergence Theorem applies to a sequence proved to be monotone and bounded. A numerical table alone establishes neither hypothesis.

This approach keeps numerical work useful without asking it to do more than it can. Computation helps identify plausible statements, possible bounds, and candidate limits. Exact arguments establish whether those statements are actually true.

Common Pitfall: Treating the Sample Maximum as a Global Bound

A frequent error is to observe that the first few terms lie below \(M\), then conclude that \(M\) is an upper bound for the sequence. That conclusion does not follow: an upper bound must work for every term.

For the sequence \(c_n\) above, the first five terms are all below \(1\), so one might incorrectly write, “The first five terms are less than \(1\); therefore \(1\) is an upper bound.” But \(c_7=2>1\). The correct statement is that the first five sampled terms are below \(1\); to claim a global upper bound, one must establish \(c_n\leq1\) for every \(n\), which this example disproves.

Rounded decimals create a related danger. A display such as \(0.9999\) may suggest that a term is less than \(1\), but calculations should be checked against exact values when possible. Rounding is useful for seeing scale and trends; it does not replace exact comparisons or proofs.

Check Your Understanding

Use the distinction between a numerical observation and a proved statement to answer the following.

  1. If the first twenty terms of a sequence increase, what can you conclude about whether the entire sequence is increasing?
  2. What does an upper bound have to satisfy that the largest term in a finite sample need not satisfy?
  3. What feature of the terms \(n/(n+1)\) suggests a possible limit of \(1\)? Why is that observation not a proof of convergence?
  4. How can two sequences agree on their first \(N\) terms but differ in convergence behaviour?
  5. When may the Monotone Convergence Theorem be used, and why is a numerical table alone insufficient?

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