Sequences and limits

A sequence is just an ordered, infinite list of numbers $a_1, a_2, a_3, \dots$ , and the first question analysis asks of any such list is blunt: does it settle down to one number as you walk further out, or not? This is the opening topic of Unit 4 ("Sequences, Series, and Power Series") in MA10001, and everything downstream — infinite series as limits of partial sums, power series convergence — reuses exactly the limit machinery built here. The goal is not the formal $\varepsilon$ - $N$ definition for its own sake but fluency with the tools that decide convergence in practice: algebraic manipulation, the algebra of limits, the squeeze theorem, and the Monotone Convergence Theorem. This article develops that toolkit and drills it on the two families of sequences — rational functions of $n$ , and factorial/exponential growth races — that show up again and again in the rest of the chapter.

The core insight

A sequence converges to $L$ if its terms get and stay arbitrarily close to $L$ as $n\to\infty$ , and in practice this is decided not by chasing the formal definition but by algebraic manipulation (divide by the dominant term), the squeeze theorem (trap a hard sequence between two easy ones), or the Monotone Convergence Theorem (bounded plus monotone forces convergence).

The core idea

A sequence $\{a_n\}_{n=1}^{\infty}$ converges to a limit $L$ , written $\lim_{n\to\infty} a_n = L$ , if for every $\varepsilon>0$ there exists an $N$ such that $|a_n - L| < \varepsilon$ for all $n>N$ — informally, the terms get and stay arbitrarily close to $L$ . If no such $L$ exists, the sequence diverges. Most limits are not computed from this definition directly; they are computed with three tools.

Algebra of limits. If $a_n\to A$ and $b_n\to B$ , then

a_n\pm b_n \to A\pm B, \qquad a_nb_n\to AB, \qquad \frac{a_n}{b_n} \to \frac{A}{B}\ (B\neq0).

where: $A$ and $B$ are the (finite) limits of the component sequences. This is the workhorse for rational expressions in $n$ : divide numerator and denominator by the highest power of $n$ present, so that every remaining piece is a constant or a term of the form $1/n^p\to 0$ , then combine.

Squeeze (sandwich) theorem. If $b_n \le a_n \le c_n$ for all sufficiently large $n$ , and $b_n\to L$ , $c_n\to L$ , then $a_n\to L$ . This is the tool of choice whenever $a_n$ is awkward to evaluate directly — typically because it involves a factorial, a bounded oscillating factor, or an alternating sign — but can be trapped between two simple sequences that both converge to the same place.

Monotone Convergence Theorem (MCT). A sequence that is monotonically increasing and bounded above converges, to its supremum; one that is monotonically decreasing and bounded below converges, to its infimum. A sequence is bounded if $|a_n|\le M$ for some fixed $M$ and all $n$ . Every convergent sequence is bounded, but the converse fails: $a_n=(-1)^n$ is bounded yet oscillates between $-1$ and $1$ forever and never converges — boundedness alone is not enough; monotonicity is what closes the gap. A subsequence of $\{a_n\}$ keeps infinitely many terms in their original order, e.g. $a_2,a_4,a_6,\dots$ . If $a_n\to L$ , every subsequence also converges to $L$ ; conversely, if two subsequences of the same sequence converge to different limits, the original sequence must diverge. This gives a quick divergence proof for oscillators: for $a_n=(-1)^n$ , the even-indexed subsequence converges to $1$ and the odd-indexed one to $-1$ , so $a_n$ itself has no limit. Four standard limits are worth memorizing outright, since they recur inside larger computations throughout the chapter:

\lim_{n\to\infty}\frac{1}{n^p}=0\ (p>0), \qquad \lim_{n\to\infty} r^n = 0\ (|r|<1), \qquad \lim_{n\to\infty} n^{1/n} = 1, \qquad \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e.

Figure

Two sequences plotted against their index n, one settling toward a horizontal limit line as n grows and the other oscillating between two fixed bounds without ever settling

Derivation ALimit of a ratio of polynomials in n

The goal is to establish, once and for all, how to evaluate $\lim_{n\to\infty} a_n$ for any sequence of the form $a_n = (p_kn^k + \cdots + p_0)/(q_mn^m+\cdots+q_0)$ with $p_k,q_m\neq0$ , since almost every rational-sequence limit in this course reduces to exactly this pattern, and doing it by dividing through by the dominant power rather than guessing from inspection is what an examiner rewards.

Step 1. Identify the dominant power in the denominator, $n^m$ , and divide every term of both numerator and denominator by it. This does not change the value of $a_n$ , since we are multiplying by $1$ in the form $(1/n^m)/(1/n^m)$ , but it rewrites $a_n$ so that each surviving piece is either a constant or a negative power of $n$ .

Step 2. After dividing, the numerator becomes $p_kn^{k-m} + p_{k-1}n^{k-1-m} + \cdots$ and the denominator becomes $q_m + q_{m-1}/n + \cdots + q_0/n^m$ . Every term in the denominator with a negative power of $n$ tends to $0$ by the standard limit $1/n^p\to0$ $(p>0)$ , so the denominator tends to the finite nonzero constant $q_m$ .

Step 3. The numerator's behaviour depends on comparing $k$ and $m$ . If $k<m$ , every surviving power of $n$ in the numerator is negative, so the numerator tends to $0$ and $a_n\to 0/q_m=0$ . If $k=m$ , the numerator tends to the constant $p_k$ , so by the algebra of limits $a_n\to p_k/q_m$ — the ratio of leading coefficients. If $k>m$ , the numerator has a positive power of $n$ surviving and grows without bound while the denominator tends to a finite constant, so $a_n$ diverges (to $+\infty$ or $-\infty$ , or oscillates in sign if the surviving power is not the same sign throughout).

Step 4. Apply this to the concrete case $a_n = (3n^2+5n-1)/(7n^2-2n+4)$ : here $k=m=2$ , so by Step 3 the limit is the ratio of leading coefficients, $3/7$ , without needing to redo the division from scratch — this is exactly the shortcut the general result buys.

Worked example

Evaluate $\displaystyle\lim_{n\to\infty} a_n$ where $a_n = \dfrac{3n^2 + 5n - 1}{7n^2 - 2n + 4}$ , and separately show that $b_n = \dfrac{n!}{n^n}$ converges to $0$ .

Step 1. For $a_n$ , both numerator and denominator have degree $k=m=2$ , so by the derivation above the limit is the ratio of leading coefficients directly — no need to divide term by term. Divide through by $n^2$ to see it explicitly:

a_n = \frac{3 + \dfrac{5}{n} - \dfrac{1}{n^2}}{7 - \dfrac{2}{n} + \dfrac{4}{n^2}} \ \longrightarrow\ \frac{3+0-0}{7-0+0} = \frac{3}{7}.

Step 2. For $b_n=n!/n^n$ , write out the factorial explicitly against $n$ copies of $n$ in the denominator:

b_n = \frac{n!}{n^n} = \frac{1}{n}\cdot\frac{2}{n}\cdot\frac{3}{n}\cdots\frac{n}{n}.

Step 3. Every factor from the second onward is at most $1$ (each is $k/n\le1$ for $k\le n$ ), and the first factor is exactly $1/n$ , so

0 \le b_n \le \frac{1}{n}.

Step 4. Since $1/n\to0$ , the squeeze theorem forces $b_n\to0$ as well. Answer: $a_n \to 3/7$ and $b_n \to 0$ .

Competitive stretch

Show $c_n = \left(1+\dfrac{2}{n}\right)^n$ converges and find its limit. Write $c_n = \left[\left(1+\frac{2}{n}\right)^{n/2}\right]^2$ ; substituting $m=n/2$ so that $2/n = 1/m$ , the inner bracket is exactly the standard limit $\left(1+1/m\right)^m\to e$ . By continuity of squaring (algebra of limits), $c_n \to e^2$ . Answer: $c_n \to e^2$ .

Common traps

Confusing "bounded" with "convergent." $a_n=(-1)^n$ is bounded between $-1$ and $1$ but never settles — it diverges by oscillation. Boundedness is necessary for convergence, never sufficient.
Forgetting the monotonicity half of MCT. The Monotone Convergence Theorem needs both boundedness and monotonicity together; a bounded but non-monotone sequence can still diverge or oscillate.
Cancelling dominant terms by eye. Students often assert the answer to an $\infty/\infty$ limit without dividing through explicitly; always factor out the highest power as shown in the derivation, rather than guessing.
Applying L'Hôpital's rule to a sequence directly. L'Hôpital is a statement about differentiable functions of a continuous variable. For $a_n=f(n)$ , one must pass to $f(x)$ , apply L'Hôpital in $x$ , then conclude about the sequence — a sequence itself cannot be differentiated.

Ask the tutor: Want to walk through more $\infty/\infty$ rational-sequence limits or practice spotting when to reach for the squeeze theorem versus the algebra of limits? Ask the AI tutor for extra worked problems.

Why isn't boundedness enough to guarantee a sequence converges?

Boundedness only says the terms never wander outside some fixed range — it says nothing about whether they stop moving around inside that range. The classic counterexample is $a_n=(-1)^n$ : it stays trapped between $-1$ and $1$ forever, satisfying the boundedness condition, but it keeps flipping between $-1$ and $1$ and never gets close to a single value. Convergence needs the terms to eventually stay close to one specific number, which is a stronger, separate requirement. That is exactly why the Monotone Convergence Theorem bundles boundedness together with monotonicity: monotonicity is what rules out this kind of endless oscillation.

When should I use the squeeze theorem instead of the algebra of limits?

Reach for the algebra of limits first when your sequence is built from sums, products, or quotients of pieces whose individual limits you already know — it is fast and direct. Switch to the squeeze theorem when the sequence itself is awkward to split apart cleanly, typically because it involves a factorial, a product of many shrinking factors, or a bounded oscillating piece (like $\sin n$ or $(-1)^n$ ) multiplying something that shrinks to zero. In those cases, find two simpler sequences that trap yours from below and above and that both converge to the same limit; the squeeze theorem then forces your sequence to the same place without your ever evaluating it directly.

Can I just use L'Hôpital's rule on sequence limits like I do for function limits?

Not directly. L'Hôpital's rule is proved for differentiable functions of a continuous real variable, and a sequence $a_n$ is only defined at integer values of $n$ — it has no derivative in the calculus sense. The correct workaround is to replace $n$ by a continuous variable $x$ , form the corresponding function $f(x)$ with $f(n)=a_n$ , apply L'Hôpital to $f(x)$ as $x\to\infty$ , and then conclude that $a_n$ has the same limit because it is a "sample" of $f$ at integer points. Skipping this step and differentiating $a_n$ "as if" it were a function of $n$ is a common but invalid shortcut.

Sequences and limits

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads