Sequences and Their Limits

Infinite lists of numbers indexed by $N$ , what it means for one to converge, the $ε$ – $N$ definition that makes the claim checkable, and the monotone convergence theorem that establishes a limit exists without naming it.

Definition

A sequence is a function from the positive integers to the reals, written $(a_{n})$ or $a_{1}, a_{2}, a_{3}, \dots$ rather than $a (n)$ . It may be given by a formula in $n$ , such as $a_{n} = \frac{3 n + 1}{n + 2}$ , or recursively, by a first term together with a rule producing each term from its predecessors.

Convergence. The sequence $(a_{n})$ converges to $L$ , written $a_{n} \to L$ or $lim_{n \to \infty} a_{n} = L$ , when for every $ε > 0$ there exists $N$ such that

n > N ⟹ | a_{n} - L | < ε .

A sequence that converges to no limit diverges. The definition is a challenge and a response: however small a tolerance $ε$ is demanded, some point $N$ in the sequence exists beyond which every term is within that tolerance of $L$ .

The quantifiers cannot be exchanged. $N$ is chosen after $ε$ and generally depends on it; a single $N$ working for all $ε$ would force the terms to equal $L$ exactly.

Bounded and monotone. The sequence is bounded if some $M$ has $| a_{n} | \leq M$ for all $n$ ; increasing if $a_{n + 1} \geq a_{n}$ for all $n$ , decreasing if $a_{n + 1} \leq a_{n}$ , and monotone if either. These are properties of the whole sequence, checkable without knowing any limit.

Monotone convergence theorem. A monotone bounded sequence converges. An increasing sequence bounded above converges to its least upper bound; a decreasing sequence bounded below converges to its greatest lower bound.

This is the unit's most useful result, because it establishes that a limit exists without producing its value, which is what makes recursively defined sequences tractable. Its converse fails: a convergent sequence is necessarily bounded but need not be monotone.

Limit laws. If $a_{n} \to L$ and $b_{n} \to M$ then $a_{n} + b_{n} \to L + M$ , $a_{n} b_{n} \to L M$ , and $a_{n} / b_{n} \to L / M$ when $M \neq 0$ . If $a_{n} \leq c_{n} \leq b_{n}$ for all large $n$ and both $a_{n}, b_{n} \to L$ , then $c_{n} \to L$ . The squeeze theorem.

Sequences and series. The partial sums $s_{n} = \sum_{k = 1}^{n} a_{k}$ of a series form a sequence, and the series converges exactly when that sequence does. So every statement about series convergence is a statement about a sequence, and the two notions must not be confused: for the harmonic series, the terms $1 / n$ converge to $0$ while the partial sums $H_{n}$ diverge.

Assumptions and scope

In the $ε$ – $N$ definition, $N$ is chosen after $ε$ and generally depends on it. A single $N$ serving every $ε$ would force $a_{n} = L$ exactly for large $n$ .
Convergence concerns the tail. Changing finitely many terms changes no limit, so no finite computation can establish or refute convergence.
The monotone convergence theorem requires both hypotheses. $(- 1)^{n}$ is bounded and diverges; $a_{n} = n$ is increasing and diverges.
Every convergent sequence is bounded, but the converse fails, so boundedness is necessary, not sufficient.
The limit laws require both limits to exist. From $a_{n} + b_{n}$ converging nothing follows about $a_{n}$ alone, as $a_{n} = n$ and $b_{n} = - n$ show.
A sequence's terms converging to zero says nothing about whether the associated series converges: $1 / n \to 0$ while $\sum 1 / n$ diverges.

Worked material

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.

Common errors

Common misconception

A bounded sequence must converge, since its terms cannot escape to infinity and so have nowhere to go but toward a limit.

Sequences and Their Limits

Assumptions and scope

Worked material

Four limits, four different arguments

Bounded is not enough

Common errors

Related units

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