Sequences and Their Limits

Orientation

The notion every convergence claim rests on

A sequence is an infinite list of numbers, $a_{1}, a_{2}, a_{3}, \dots$ , one for each positive integer. That is all: a function on $N$ , written with a subscript instead of brackets.

The only question usually worth asking is where the list is heading. And it is a subtler question than it looks, because a sequence can approach a value without ever attaining it:

a_{n} = \frac{3 n + 1}{n + 2} : 1.333, 1.75, 2.286, 2.583, \dots, 2.951, \dots, 2.9995, \dots

No term equals 3, yet 3 is plainly the answer. "Heading toward $L$ " cannot mean "eventually equals $L$ ", so it needs a definition that does not mention arriving.

The definition is the $ε$ – $N$ one, and it works by turning the claim into a challenge. Name any tolerance $ε$ , however small, and the claim survives if some position $N$ can be produced beyond which every term is within $ε$ of $L$ . For the sequence above the error is exactly $5 / (n + 2)$ , so a tolerance of $0.001$ is met from $n = 4999$ onward. That is what convergence means.

Why this unit sits underneath others. The series unit defines an infinite sum as the limit of its partial sums $s_{n} = a_{1} + \dots + a_{n}$ , and those form a sequence. So every convergence test there is a statement about a sequence, and the distinction the series unit most needs is one only sequences can state: for the harmonic series the terms $1 / n$ converge to zero while the partial sums grow without bound, passing 12 by $n = 100,000$ and never stopping.

The numerical unit leaned on the same idea from the other side. Its refinement check, running at $h$ and again at $h / 2$ to see whether the answers agree, is an appeal to the convergence of a computed sequence.

The hard case, and the unit's best result. The $ε$ – $N$ definition requires knowing $L$ in advance, which is useless for a sequence defined by a rule such as

a_{1} = 1, a_{n + 1} = \sqrt{2 + a_{n}},

where no formula for $a_{n}$ exists. The monotone convergence theorem handles it: the terms increase and never reach 2, so a limit must exist, and only then can the rule be solved to find it is exactly 2. The theorem establishes that a limit exists; the rule then determines its value.

Definition

Reading the quantifiers

Almost every misunderstanding of convergence is a misreading of the quantifier order.

The statement. $a_{n} \to L$ means

\forall ε > 0 \exists N \forall n > N : | a_{n} - L | < ε .

Read left to right: given any tolerance, there is a position, after which every term meets it.

Why the order cannot be exchanged. Putting $\exists N$ first would demand a single $N$ working for every $ε$ simultaneously, so $| a_{n} - L |$ would be smaller than every positive number for $n > N$ , forcing $a_{n} = L$ exactly. That is a much stronger and almost always false condition. $N$ depends on $ε$ , and for $a_{n} = \frac{3 n + 1}{n + 2}$ it grows as $ε$ shrinks: $ε = 0.1$ needs $n > 48$ , $ε = 0.01$ needs $n > 498$ , $ε = 0.001$ needs $n > 4998$ .

Why "for all $n > N$ " and not "for some". Convergence requires the terms to stay close, not merely to come close occasionally. The sequence $0, 1, 0, \frac{1}{2}, 0, \frac{1}{3}, \dots$ returns to within any tolerance of $0$ infinitely often yet does not converge, because it keeps leaving again.

What $| a_{n} - L |$ measures. The distance from the term to the candidate limit. The definition says nothing about consecutive terms approaching each other, and nothing about the distance decreasing at every step. A convergent sequence may move further away at some steps, provided it eventually stays inside every tolerance.

Convergence is a property of the tail. Only $n > N$ appears, so changing finitely many terms cannot affect any limit. Two consequences follow: no computation of finitely many terms can prove convergence, and a sequence that looks settled for a million terms may still diverge.

Boundedness and monotonicity are different in kind. Both are checkable without knowing a limit, that is exactly what makes the monotone convergence theorem useful. Note also that monotone means always increasing or always decreasing; a sequence that rises then falls is neither, whatever its tail does.

What the theorem does and does not give. It asserts that the limit exists; it does not say what the limit is. An increasing sequence bounded above converges to its least upper bound, which is a characterisation rather than a computation, and finding the value normally needs a separate argument, for a recursive sequence, solving the fixed-point equation.

Both hypotheses are needed, and dropping either is fatal in a different way: $a_{n} = n$ is increasing and unbounded, so it diverges to infinity; $a_{n} = (- 1)^{n}$ is bounded and not monotone, and diverges by oscillation.

Intuition

Convergence as a game with a challenger

Played out on a concrete sequence, the challenge-and-response structure becomes visible.

The sequence. $a_{n} = \frac{3 n + 1}{n + 2}$ , claimed limit $L = 3$ .

The error, exactly. Rather than inspecting terms, compute the distance:

| a_{n} - 3 | = | \frac{3 n + 1}{n + 2} - 3 | = | \frac{3 n + 1 - 3 n - 6}{n + 2} | = \frac{5}{n + 2} .

One formula now answers every challenge.

The game.

Challenger demands	Response: solve $\frac{5}{n + 2} < ε$	Take $N =$	Check at $n = N + 1$
$ε = 0.1$	$n > 48$	$48$	$a_{49} = 2.90196078$ , error $0.098$
$ε = 0.01$	$n > 498$	$498$	$a_{499} = 2.99001996$ , error $0.00998$
$ε = 0.001$	$n > 4998$	$4998$	$a_{4999} = 2.99900020$ , error $0.00099$

Each response is produced by solving, not guessing: $\frac{5}{n + 2} < ε ⟺ n > \frac{5}{ε} - 2$ . Because that formula answers any $ε$ , the challenger can never win, and that is precisely what convergence asserts.

Why the definition avoids motion. It is tempting to say the terms "get closer and closer", but that phrasing over-promises. Convergence does not require the distance to shrink at every step, only that it eventually stay below each tolerance. Consider

1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots with every tenth term replaced by something slightly larger.

Such a sequence still converges to $0$ while repeatedly stepping backwards. The $ε$ – $N$ formulation says nothing about individual steps, which is why it survives these cases.

Where the intuition must be surrendered. The whole picture above assumed $L$ was known, $3$ came from the algebra before the game began. For

a_{1} = 1, a_{n + 1} = \sqrt{2 + a_{n}},

there is no formula for $a_{n}$ and therefore no $L$ to challenge with. Computing terms gives $1$ , $1.732051$ , $1.931852$ , $1.982890$ , $1.995718$ , $1.998929$ , suggestive, but a finite list can never establish a limit, since convergence is a property of the tail.

The monotone convergence theorem is what rescues the situation, and its logic runs in the opposite direction to the game: instead of verifying closeness to a known target, it argues from the shape of the sequence that a target must exist. Increasing and capped leaves nowhere else to go. Only afterwards is the value extracted, by solving $L = \sqrt{2 + L}$ .

Simulation

A convergent sequence, its epsilon band, and the index N

The sequence, the band 3 ± ε, and the integer N from which the tail stays inside

The sequence, the band 3 ± ε, and the integer N from which the tail stays inside

A scatter of the sequence $a_{n} = (3 n + 1) / (n + 2)$ for $n$ from 1 to 120, rising from about 1.33 toward a horizontal grey line at $L = 3$ which it never reaches. Two red horizontal lines at $3 - ε$ and $3 + ε$ bound a band whose width the reader sets, and a green vertical rule stands at the integer index $N$ . The exact relation is $| a_{n} - 3 | = 5 / (n + 2)$ , so the band is entered when $n$ exceeds the real threshold $T = 5 / ε - 2$ , and the index from the definition of convergence is the first integer past it, $N = ⌊ 5 / ε - 2 ⌋ + 1$ . Every plotted point from $n = N$ onward lies inside the band and the point at $n = N - 1$ lies outside it. As $ε$ decreases the band narrows and the rule moves right: $ε = 0.5$ gives $N = 9$ , $ε = 0.1$ gives $N = 49$ , and $ε = 0.05$ gives $N = 99$ , still inside the plotted range.

The points are $a_{n} = \frac{3 n + 1}{n + 2}$ and the grey line is the claimed limit $L = 3$ . Set $ε$ and the red lines become the band $3 \pm ε$ ; the green rule stands at the index $N$ from which every term is inside.

Convergence is not that the terms reach $L$ — none of them does — but that only finitely many stay outside any band you name. Since $| a_{n} - 3 | = \frac{5}{n + 2}$ , the condition $| a_{n} - 3 | < ε$ holds exactly when

n > T (ε) = \frac{5}{ε} - 2.

$T$ is a real threshold, not an index. The $N$ in the definition is a natural number, so it is the first integer past $T$ :

N (ε) = ⌊ \frac{5}{ε} - 2 ⌋ + 1.

At $ε = 0.5$ that is $N = 9$ ; at $ε = 0.1$ , $N = 49$ ; at the smallest setting $ε = 0.05$ , $N = 99$ , which is why the plot runs to $120$ . The green rule draws $N$ , and you can check it: the term at $n = N$ is inside the band and the one at $n = N - 1$ is not.

Watch the two move together: shrink $ε$ and the band narrows while the rule slides right. That pairing — for each $ε$ there is an $N$ — is the definition, and the quantifier order is the content: $N$ depends on $ε$ , and does.

Example

Four limits, four different arguments

Each of these converges, and each needs a different reason.

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1. A rational formula: divide by the dominant power.

a_{n} = \frac{3 n + 1}{n + 2} .

Dividing numerator and denominator by $n$ gives $\frac{3 + 1 / n}{1 + 2 / n}$ . Since $1 / n \to 0$ , the limit laws give $\frac{3 + 0}{1 + 0} = 3$ .

Confirmation. $a_{100} = 2.9509803922$ , $a_{1000} = 2.9950099800$ , $a_{10000} = 2.9995001000$ , with errors $4.9 \times 10^{- 2}$ , $5.0 \times 10^{- 3}$ , $5.0 \times 10^{- 4}$ , falling by a factor of 10 per decade, exactly as $5 / (n + 2)$ predicts.

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2. Exponential beats polynomial.

a_{n} = \frac{n}{2^{n}} .

Both parts grow, so the limit laws do not apply directly and the question is which wins. The terms: $0.5$ , $0.5$ , $0.375$ , $0.15625$ at $n = 1, 2, 3, 5$ , then $0.009765625$ at $n = 10$ , $1.907 \times 10^{- 5}$ at $n = 20$ , $2.794 \times 10^{- 8}$ at $n = 30$ , $4.441 \times 10^{- 14}$ at $n = 50$ .

The limit is $0$ . Note the sequence is not decreasing at the first step, $a_{1} = a_{2} = 0.5$ , which costs nothing, since convergence is a property of the tail.

Why exponential wins. The ratio of consecutive terms is $\frac{a_{n + 1}}{a_{n}} = \frac{n + 1}{2 n} \to \frac{1}{2}$ , so eventually each term is little more than half its predecessor, and a sequence shrinking geometrically tends to zero however large it starts.

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3. Factorial beats exponential.

a_{n} = \frac{2^{n}}{n!} .

Terms: $2$ , $2$ , $0.2666667$ , $2.822 \times 10^{- 4}$ , $2.506 \times 10^{- 8}$ , $4.310 \times 10^{- 13}$ at $n = 1, 2, 5, 10, 15, 20$ . The limit is $0$ , and the collapse is far faster than in the previous example.

Why. The ratio is $\frac{a_{n + 1}}{a_{n}} = \frac{2}{n + 1}$ , which tends to $0$ rather than to a constant, so the shrinkage accelerates. This ranking, factorial over exponential over polynomial, is what makes the ratio test work in the series unit.

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4. The squeeze theorem, where the terms do not settle down.

a_{n} = \frac{\sin n}{n} .

Here $\sin n$ oscillates forever and has no limit, so no dominant-term argument applies. But $| \sin n | \leq 1$ gives

- \frac{1}{n} \leq \frac{\sin n}{n} \leq \frac{1}{n},

and both bounds tend to $0$ , so the middle is squeezed to $0$ .

Confirmation. $a_{1} = + 0.8414709848$ , $a_{10} = - 0.0544021111$ , $a_{100} = - 0.0050636564$ , $a_{1000} = + 0.0008268795$ , inside $\pm 1 / n$ at every point, and changing sign repeatedly while converging.

This last is the one to remember against the intuition that a convergent sequence approaches its limit steadily from one side.

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5. A limit that defines a constant.

a_{n} = {(1 + \frac{1}{n})}^{n} .

The base tends to 1 and the exponent to infinity, so the limit laws give nothing: $1^{\infty}$ is indeterminate, as the l'Hôpital unit explains. The terms:

$n$	$a_{n}$	$e - a_{n}$
$1$	$2.000000000000$	$7.183 \times 10^{- 1}$
$10$	$2.593742460100$	$1.245 \times 10^{- 1}$
$100$	$2.704813829422$	$1.347 \times 10^{- 2}$
$1000$	$2.716923932236$	$1.358 \times 10^{- 3}$
$10^{4}$	$2.718145926825$	$1.359 \times 10^{- 4}$
$10^{5}$	$2.718268237192$	$1.359 \times 10^{- 5}$
$10^{6}$	$2.718280469096$	$1.359 \times 10^{- 6}$

The limit is $e = 2.718281828459$ . The errors fall by exactly a factor of 10 per decade of $n$ , settling to $1.359 / n$ , slow convergence, and a poor way to compute $e$ , but this limit is one standard definition of it.

Non-example

Bounded is not enough

The claim to be refuted. A bounded sequence cannot run off to infinity, so surely it must settle somewhere.

The counterexample.

a_{n} = (- 1)^{n} : - 1, 1, - 1, 1, - 1, 1, - 1, 1, \dots

Every term satisfies $| a_{n} | \leq 1$ , so the sequence is bounded. It does not converge.

Why not, by the definition. Suppose $a_{n} \to L$ . Take the challenge $ε = 1$ . Convergence would give an $N$ with $| a_{n} - L | < 1$ for all $n > N$ . But beyond any $N$ the sequence takes both values $- 1$ and $1$ , so that would require

| 1 - L | < 1 and | - 1 - L | < 1,

forcing $L > 0$ and $L < 0$ simultaneously. No such $L$ exists, so the sequence converges to nothing.

What goes wrong in the intuition. "Nowhere to go but toward a limit" assumes the only alternative to converging is escaping. There is a third option: moving back and forth forever inside a bounded region. Boundedness forbids escape; it does nothing about oscillation.

The subsequence view. The even-indexed terms form the constant sequence $1, 1, 1, \dots \to 1$ , and the odd-indexed terms $\to - 1$ . A sequence converges only if every subsequence converges to the same value, so two subsequences with different limits refute convergence immediately. This is usually the quickest way to show a bounded sequence diverges.

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Two further things that are not enough.

Monotone alone is not enough. $a_{n} = n$ increases at every step and diverges to infinity. Boundedness is what rules this out, so the monotone convergence theorem needs both hypotheses, and each blocks a different failure: monotonicity blocks oscillation, boundedness blocks escape.

A finite computation is not enough. The sequence $b_{n} = \frac{n}{2^{n}}$ has $b_{1} = b_{2} = 0.5$ , so a check of the first two terms would suggest a constant sequence; it in fact tends to $0$ . More sharply, since convergence depends only on the tail, no finite number of terms can establish it: a sequence agreeing with $1 / n$ for a million terms and equal to $1$ thereafter converges to $1$ , not $0$ .

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The confusion this most often causes. In the series unit, the terms of the harmonic series satisfy $1 / n \to 0$ , a convergent and bounded sequence, while the partial sums

H_{n} = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}

diverge: $H_{10} = 2.928968$ , $H_{100} = 5.187378$ , $H_{1000} = 7.485471$ , $H_{10000} = 9.787606$ , $H_{100000} = 12.090146$ . Growth this slow looks like settling, and it is not; the check block shows why it never stops.

Two sequences are in play, the terms and the partial sums, and they behave differently. Conflating them is the error the series unit names as its own misconception, and it begins here, in reading a bounded or slowly-changing sequence as a convergent one.

Theorem

Monotone and bounded implies convergent

Theorem (monotone convergence). An increasing sequence bounded above converges, and its limit is the least upper bound of its terms. A decreasing sequence bounded below converges to the greatest lower bound.

Why it holds. Let $(a_{n})$ be increasing with $a_{n} \leq M$ for all $n$ . The set ${a_{n}}$ is nonempty and bounded above, so by the completeness of $R$ it has a least upper bound $L$ .

Given $ε > 0$ , the number $L - ε$ is not an upper bound, otherwise $L$ would not be the least, so some term satisfies $a_{N} > L - ε$ . Since the sequence increases, every $n > N$ has $a_{n} \geq a_{N} > L - ε$ , and since $L$ is an upper bound, $a_{n} \leq L$ . Hence

L - ε < a_{n} \leq L ⟹ | a_{n} - L | < ε for all n > N,

which is the definition of $a_{n} \to L$ .

The argument produces $N$ from the failure of $L - ε$ to be an upper bound, never from a formula for $a_{n}$ . That is why the theorem applies to sequences with no closed form, and it is the reason for its usefulness.

It rests on completeness: in $Q$ the theorem is false, since an increasing bounded sequence of rationals can approach an irrational and have no rational limit. The sequence $a_{n + 1} = \sqrt{2 + a_{n}}$ happens to converge to the rational 2, but $1, 1.4, 1.41, 1.414, \dots$ increasing and bounded by 2 converges to $\sqrt{2}$ , which is not in $Q$ .

Both hypotheses are needed.

Sequence	Monotone	Bounded	Converges
$a_{n + 1} = \sqrt{2 + a_{n}}$ , $a_{1} = 1$	yes	yes	yes, to 2
$a_{n} = n$	yes	no	no, diverges to $\infty$
$a_{n} = (- 1)^{n}$	no	yes	no, oscillates
$a_{n} = (- 1)^{n} / n$	no	yes	yes, to 0

The last row matters: the conditions are sufficient, not necessary. A sequence failing them may still converge, so the theorem is a tool for proving convergence and never for refuting it.

Corollary (convergent implies bounded). If $a_{n} \to L$ , take $ε = 1$ to get $N$ with $| a_{n} - L | < 1$ for $n > N$ ; then every term is bounded by the largest of $| a_{1} |, \dots, | a_{N} |$ and $| L | + 1$ . So boundedness is necessary for convergence but not sufficient, as $(- 1)^{n}$ shows, and an unbounded sequence can be dismissed at once.

Worked application, verified. For $a_{1} = 1$ , $a_{n + 1} = \sqrt{2 + a_{n}}$ :

$n$	$a_{n}$
1	$1.000000000000$
2	$1.732050807569$
3	$1.931851652578$
4	$1.982889722748$
5	$1.995717846477$
8	$1.999933067835$
11	$1.999998954179$

Every term was confirmed to exceed its predecessor and to stay below 2. The theorem therefore guarantees a limit $L$ exists; the procedure block extracts its value.

Note what would be unavailable without the theorem: the table alone shows eleven terms below 2 and proves nothing, since convergence depends on the infinite tail.

Procedure

Deciding whether a sequence converges

Input. A sequence given by a formula in $n$ or by a recursive rule.

Step 1 — Check boundedness first if divergence is plausible. An unbounded sequence cannot converge, by the corollary to the monotone convergence theorem. This disposes of $a_{n} = n$ , $a_{n} = n^{2} / (n + 1)$ and $a_{n} = 2^{n}$ immediately.

Step 2 — If there is a formula, compute the limit directly. Choose by the shape:

Shape	Method
ratio of polynomials	divide top and bottom by the highest power of $n$
polynomial against exponential, or exponential against factorial	the faster-growing denominator wins; confirm with the ratio $a_{n + 1} / a_{n}$
bounded factor over something growing	squeeze between $\pm$ the bound over that quantity
$a_{n} = f (n)$ for a function with a known limit at infinity	take the limit of $f$ ; l'Hôpital applies to $f$ , not to $(a_{n})$ directly
indeterminate powers such as $(1 + 1 / n)^{n}$	take logarithms, or recognise a standard limit

Step 3 — If the sequence is recursive, do not try for a formula. Instead:

(a) Show monotonicity. Compare $a_{n + 1}$ with $a_{n}$ , usually by induction, assuming $a_{n} > a_{n - 1}$ and deducing $a_{n + 1} > a_{n}$ from the rule.

(b) Show boundedness. Guess a bound from the first few terms and prove it by induction: if $a_{n} < M$ then the rule must give $a_{n + 1} < M$ .

(c) Cite the theorem to conclude a limit $L$ exists. The next step is invalid without it.

(d) Solve for the limit. Let $n \to \infty$ on both sides of $a_{n + 1} = g (a_{n})$ . Since $(a_{n + 1})$ is the same sequence shifted, both sides tend to $L$ , giving the fixed-point equation $L = g (L)$ .

(e) Reject inadmissible roots. The equation may have roots the sequence cannot approach: negative ones when all terms are positive, or values below an increasing sequence's first term.

Step 4 — To produce $N$ for a given $ε$ , when the limit is known: write $| a_{n} - L |$ as a single simplified expression, set it $< ε$ , and solve for $n$ . Take $N$ to be any integer at least as large as the bound obtained.

Step 5 — To show divergence, use whichever applies:

unbounded, which is enough on its own;
two subsequences with different limits, the quickest route for a bounded oscillating sequence, as with $(- 1)^{n}$ ;
direct contradiction from the definition, as in the non-example block.

Where it goes wrong.

Solving the fixed-point equation before establishing existence. The manipulation "let $n \to \infty$ on both sides" presumes a limit to pass to. Applied to $a_{n + 1} = 2 a_{n}$ with $a_{1} = 1$ it yields $L = 2 L$ , hence $L = 0$ , while the sequence diverges to infinity. The equation is only meaningful once convergence is known.
Concluding convergence from boundedness alone, or from a table of terms. Neither is sufficient, and no finite table can be.
Forgetting to reject a root. For $a_{n + 1} = \sqrt{2 + a_{n}}$ the fixed-point equation is $L^{2} - L - 2 = 0$ with roots $2$ and $- 1$ ; the terms are positive, so $- 1$ is impossible.
Applying l'Hôpital's rule to a sequence. It requires a differentiable function of a real variable; the legitimate move is to apply it to $f (x)$ and then restrict to integer $x$ .
Treating monotone as "mostly increasing". One step in the wrong direction disqualifies the theorem, though a sequence monotone from some point on is fine, since convergence depends only on the tail.

Worked example

A recursive sequence, from existence to value

Problem. Let $a_{1} = 1$ and $a_{n + 1} = \sqrt{2 + a_{n}}$ . Show the sequence converges and find its limit.

There is no formula for $a_{n}$ , so the $ε$ – $N$ definition has no target to aim at. This is the case the monotone convergence theorem exists for.

Step 0 — Compute a few terms to see what to prove.

a_{1} = 1, a_{2} = \sqrt{3} = 1.732051, a_{3} = \sqrt{3.732051} = 1.931852, a_{4} = 1.982890, a_{5} = 1.995718 .

They increase and appear to approach 2. Those are the two claims to establish; computing terms is reconnaissance, not proof.

Step 3(b) — Bounded above by 2, by induction.

Base: $a_{1} = 1 < 2$ .

Step: assume $a_{n} < 2$ . Then

a_{n + 1} = \sqrt{2 + a_{n}} < \sqrt{2 + 2} = \sqrt{4} = 2 .

So $a_{n} < 2$ for every $n$ . Notice how the bound 2 was found: it is the value that reproduces itself under the rule, which is why the induction closes exactly.

Step 3(a) — Increasing, by induction.

Base: $a_{2} = 1.732051 > 1 = a_{1}$ .

Step: assume $a_{n} > a_{n - 1}$ . Then $2 + a_{n} > 2 + a_{n - 1}$ , and the square root is increasing, so

a_{n + 1} = \sqrt{2 + a_{n}} > \sqrt{2 + a_{n - 1}} = a_{n} .

So the sequence increases at every step.

Verified numerically for the first eleven terms: each exceeds its predecessor and each stays below 2.

$n$	$a_{n}$	increasing	$< 2$
1	$1.000000000000$	—	✓
2	$1.732050807569$	✓
3	$1.931851652578$	✓
4	$1.982889722748$	✓
5	$1.995717846477$	✓
6	$1.998929174953$	✓
7	$1.999732275819$	✓
8	$1.999933067835$	✓
9	$1.999983266889$	✓
10	$1.999995816718$	✓
11	$1.999998954179$	✓

Step 3(c) — Cite the theorem. The sequence is increasing and bounded above, so a limit $L$ exists. This is the step that licenses everything following; without it, the algebra below would be manipulating a symbol that might denote nothing.

Step 3(d) — Solve the fixed-point equation. Take $n \to \infty$ in $a_{n + 1} = \sqrt{2 + a_{n}}$ . This is legitimate now that both sides have limits, and both equal $L$ because $(a_{n + 1})$ is the same sequence shifted by one place:

L = \sqrt{2 + L} .

Squaring, $L^{2} = 2 + L$ , so

L^{2} - L - 2 = 0 ⟹ (L - 2) (L + 1) = 0 ⟹ L = 2 or L = - 1 .

Step 3(e) — Reject the inadmissible root. Every term is positive, since $a_{1} = 1 > 0$ and a square root is never negative, so the limit cannot be $- 1$ . Also, the sequence increases from $a_{1} = 1$ , so $L \geq 1$ .

L = 2

Verification. The fixed point checks: $\sqrt{2 + 2} = \sqrt{4} = 2$ . And $a_{11} = 1.999998954179$ , within $1.05 \times 10^{- 6}$ of 2.

Why the order of steps matters. Reversing it, by solving $L = \sqrt{2 + L}$ first and calling $2$ the limit, is the error the procedure block warns about. Apply that reasoning to $a_{1} = 1$ , $a_{n + 1} = 2 a_{n}$ : the fixed-point equation $L = 2 L$ gives $L = 0$ , yet the sequence is $1, 2, 4, 8, \dots$ , diverging to infinity. The algebra is identical; what differs is that the second sequence is unbounded, so no limit exists to substitute. The equation says what the limit must be if there is one, never that there is one.

Check your understanding

A convergent sequence with a divergent series

The harmonic series is where the distinction between a sequence and its partial sums becomes unavoidable. Work through it deliberately, because it is the case that decides whether the definition has been understood.

The terms. $a_{n} = \frac{1}{n}$ : $1, 0.5, 0.333, 0.25, \dots$ This sequence converges to 0. Given $ε$ , take $N > 1 / ε$ and every later term satisfies $| 1 / n - 0 | < ε$ . Straightforward.

The partial sums. $H_{n} = 1 + \frac{1}{2} + \dots + \frac{1}{n}$ . This is a different sequence, built from the first, and the question of its convergence is the question of whether $\sum 1 / n$ converges.

What the numbers suggest, and why they mislead.

$n$	$H_{n}$
$1$	$1.000000$
$10$	$2.928968$
$100$	$5.187378$
$1000$	$7.485471$
$10^{4}$	$9.787606$
$10^{5}$	$12.090146$

Growth is slow, a hundredfold increase in $n$ adds about $4.6$ , and the increments are shrinking, so the sums look as though they might be settling. They are not.

Why they never settle. Group the terms in blocks of doubling length:

\underset{\geq 1 / 2}{\underset{⏟}{\frac{1}{2}}} + \underset{> \frac{1}{4} + \frac{1}{4} = \frac{1}{2}}{\underset{⏟}{\frac{1}{3} + \frac{1}{4}}} + \underset{> 4 \times \frac{1}{8} = \frac{1}{2}}{\underset{⏟}{\frac{1}{5} + \dots + \frac{1}{8}}} + \underset{> 8 \times \frac{1}{16} = \frac{1}{2}}{\underset{⏟}{\frac{1}{9} + \dots + \frac{1}{16}}} + \dots

Each block contributes more than $\frac{1}{2}$ , because it contains enough terms, each at least as large as its smallest member. So

H_{2 n} - H_{n} \geq \frac{1}{2} for every n .

Verified:

$n$	$H_{2 n} - H_{n}$	$\geq \frac{1}{2}$
$1$	$0.500000$	✓
$2$	$0.583333$	✓
$4$	$0.634524$	✓
$8$	$0.662872$	✓
$16$	$0.677766$	✓

Since every doubling adds at least $\frac{1}{2}$ , the sums exceed any bound: after $k$ doublings $H_{2^{k}} \geq 1 + k / 2$ , which grows without limit. $(H_{n})$ diverges.

This also settles it by the definition: a convergent sequence must satisfy $| H_{m} - H_{n} | < ε$ for all large $m, n$ , and taking $ε = \frac{1}{4}$ with $m = 2 n$ contradicts the inequality above at every $n$ .

The comparison.

	Terms $1 / n$	Partial sums $H_{n}$
Monotone	decreasing	increasing
Bounded	yes, by 1	no
Converges	yes, to 0	no

Both are monotone. They differ in boundedness, which is what decides convergence. The theorem applies to the terms and not to the sums.

How slowly. $H_{n}$ tracks $\ln n + γ$ with $γ = 0.5772156649$ closely: at $n = 100$ the sum is $5.187378$ against $5.182386$ ; at $n = 10^{5}$ , $12.090146$ against $12.090141$ . Since $\ln n \to \infty$ , the growth is genuine but logarithmic, reaching $H_{n} > 100$ needs about $e^{99.42} \approx 10^{43}$ terms. A sequence can diverge far too slowly for any computation to reveal it, which is the practical reason the grouping argument is needed.

The point to carry forward. "The terms go to zero" and "the series converges" are claims about two different sequences. The first is necessary for the second and nowhere near sufficient, which is exactly what the series unit's $n$ th-term test says, and why that test can only ever prove divergence.

Application

Iteration: where recursive sequences come from

Recursively defined sequences are not a textbook curiosity. They are what an iterative algorithm produces, and the question of whether such a sequence converges is the question of whether the algorithm works.

Newton's method. To solve $f (x) = 0$ , iterate

x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})},

which is a recursive sequence exactly like the one in the worked example. Applied to $f (x) = x^{2} - 2$ from $x_{1} = 2$ it gives $\sqrt{2}$ , and the fixed-point equation $L = L - \frac{L^{2} - 2}{2 L}$ reduces to $L^{2} = 2$ , so if it converges, it converges to a root. That conditional is the whole issue: Newton's method does not always converge, and establishing that it does, for a given $f$ and starting point, is the same monotone-and-bounded style of argument used for $a_{n + 1} = \sqrt{2 + a_{n}}$ .

Fixed-point iteration generally. Any equation rearranged as $x = g (x)$ suggests the iteration $x_{n + 1} = g (x_{n})$ . The sequence converges when $g$ contracts distances near the fixed point, and it diverges when $g$ expands them, which is why the same equation, rearranged differently, can give an iteration that works and one that does not. The fixed-point equation cannot tell the two apart: both rearrangements have the same solution, and only the sequence's behaviour differs.

Numerical solutions of differential equations. Every method in the previous unit generates a sequence $y_{0}, y_{1}, y_{2}, \dots$ by a recursive rule, $y_{n + 1} = y_{n} + h f (t_{n}, y_{n})$ for Euler. Two convergence questions arise, and they are different:

Within one run, does the computed sequence stay bounded? That is the stability question, and $y^{'} = - 20 y$ with $h = 0.2$ answered it with $1, - 3, 9, - 27, 81, - 243$ . An unbounded sequence, diverging by the criterion of this unit.
Across runs, do the answers at $h, h / 2, h / 4, \dots$ converge as $h \to 0$ ? That is the accuracy question, and the refinement check compares successive terms of that sequence of answers.

Compound interest and the definition of $e$ . Interest compounded $n$ times a year at rate 100% turns 1 unit into $(1 + 1 / n)^{n}$ . Compounding more often gives $2$ , $2.593742$ , $2.704814$ , $2.716924$ , $2.718146$ at $n = 1, 10, 100, 1000, 10^{4}$ , increasing, bounded, and converging to $e = 2.718281828459$ . Continuous compounding is a statement about a sequence's limit, and the monotone convergence theorem is what says the limit exists.

Long-run behaviour of dynamical systems. A population model $p_{n + 1} = g (p_{n})$ tracks numbers year by year. Whether the population settles at an equilibrium, oscillates between values, or grows without bound is exactly whether the sequence converges, oscillates, or is unbounded. The three behaviours this unit distinguishes. Equilibria are fixed points of $g$ , and the same caution applies: a fixed point exists as a solution of an equation whether or not any trajectory approaches it.

---

In every case the sequence is generated by a rule rather than a formula, so the terms can be computed indefinitely but no closed form is available. The recurring need is to decide, in advance of computing, whether the iteration will settle, because a computation cannot answer it. A billion terms neither prove convergence, since it depends on the tail, nor rule it out, since $H_{n}$ shows divergence can be slower than any feasible computation.

That is why the monotone convergence theorem earns its place: monotonicity and boundedness are checkable from the rule itself, and together they settle a question no amount of arithmetic can.

Sequences and Their Limits

The notion every convergence claim rests on

Reading the quantifiers

Convergence as a game with a challenger

A convergent sequence, its epsilon band, and the index N

Four limits, four different arguments

Bounded is not enough

Monotone and bounded implies convergent

Deciding whether a sequence converges

A recursive sequence, from existence to value

A convergent sequence with a divergent series

Iteration: where recursive sequences come from

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