Sequences in Calculus 1: limits and tests for the midterm

End of September 2026, first year of engineering, maths, physics or economics: in many Calculus 1 (Analisi Matematica 1) courses, sequences arrive in October, straight after the real numbers and the supremum. It is the first topic of the course that is really built on limits, and it is usually also the first topic of the first midterm (prova in itinere), which at many universities falls in mid-November, about seven weeks from now. Calendars vary a lot from course to course, so the real point of reference remains your exam syllabus, not a “universal” date.

Sequences share the vocabulary of limits of functions. If you still lack that, limits explained properly covers the common part (definition, indeterminate forms, L’Hôpital) and I don’t repeat it here. What changes with sequences is the domain: no longer a continuous interval of real numbers, but only the natural numbers. This difference, which looks like a detail, generates some tools that functions don’t need (ratio test, root test, hierarchy of infinities), and it is exactly what the midterm asks for.

What a sequence is, what its limit is

A sequence is a function that associates with each natural number $n$ a real number $a_n$: $a:\mathbb{N}\to \mathbb{R}$. It is written $\{a_n{\}}_{n\in \mathbb{N}}$ or simply $a_n$. Unlike a function of a real variable, the domain is discrete: it makes sense to ask what happens as $n\to +\infty$ (the only sensible limit for a sequence, since $n$ always grows in steps of 1), not as $n$ tends to a finite value.

The limit of a sequence behaves as you expect from limits of functions, with three possible outcomes:

• Convergent: ${\lim }_{n\to \infty }{a}_n=L$, a finite real number. Example: $a_n=\frac{1}{n}\to 0$.
• Divergent: ${\lim }_{n\to \infty }{a}_n=+\infty$ or $-\infty$. Example: $a_n=n^2\to +\infty$.
• Irregular (or indeterminate): the limit does not exist, neither finite nor infinite. The example to always keep in mind is

which is alternately $1,-1,1,-1,\dots$ without ever settling or blowing up. It is not $0$ (the average), it is not $\pm 1$: it does not exist, and that is as legitimate an outcome as the other two, not a calculation error.

Monotone sequences: when the limit always exists

A sequence is monotonically increasing if $a_{n+1}\ge a_n$ for every $n$ (decreasing if the opposite holds). For monotone sequences a theorem holds that generic functions don’t have:

Existence theorem for the limit of monotone sequences. Every monotone sequence has a limit: finite if the sequence is bounded, $+\infty$ (or $-\infty$) if it is not.

This is a strong result: a monotone sequence can never be irregular. The case $(-1{)}^n$ seen above is not monotone (it goes up and down continually), and that is exactly why it can afford to be irregular.

The number $e$ as a standard limit

The theorem we have just seen is not only theory: it is what guarantees the existence of one of the most important limits in all of calculus, Euler’s number:

It can be proved that this sequence is increasing and bounded above, so by the theorem above it converges to a finite value, which by definition we call $e\approx 2.71828\dots$. You don’t need to redo that proof in every exercise: the standard limit is used directly, and its generalised form

comes up often in the midterm, as in the third exercise below.

Algebra of limits and indeterminate forms

Sum, product and quotient of limits of convergent sequences behave exactly as for functions, and the same seven indeterminate forms hold here:

There is, though, an important practical difference. L’Hôpital and Taylor expansions are tools defined for functions of a real variable, with a derivative with respect to a continuous variable: they make no sense applied directly to $n$, which jumps from integer to integer. To use them on a sequence you must first move to the corresponding function $f(x)$, obtained by writing the same expression but letting $x$ vary continuously, compute ${\lim }_{x\to \infty }f(x)$ with the tools for functions, and only then conclude that ${\lim }_{n\to \infty }f(n)$ has the same limit, because if a function has a limit as $x\to \infty$, the sequence of its values at the integers has the same limit too. The reverse step, however, does not hold: a sequence can converge even when the corresponding function has no limit at all. The example to remember is $a_n=\sin (\pi n)$: it is $0$ for every integer $n$ (a constant sequence, trivially converging to $0$), whereas $f(x)=\sin (\pi x)$ as a real function oscillates forever and has no limit as $x\to \infty$. Confusing the two, treating a sequence as if it were “the same thing” as the function that generates it, is one of the most common mistakes, taken up again below.

The hierarchy of infinities

When logarithms, powers, exponentials and factorials appear together in an indeterminate form, it is worth knowing the order of growth by heart, from slowest to fastest, as $n\to +\infty$:

The symbol $\ll$ is read “is negligible compared with”: it means that the ratio of the term on the left to the one on the right tends to $0$. This hierarchy resolves a great many $\frac{\infty }{\infty }$ forms instantly, with no other tools: in a ratio between a factorial and a power, for example, the factorial always wins, whatever the base of the power.

Ratio test and root test

For sequences with positive terms involving factorials or powers, two tests spare you from having to reason “by eye” about the hierarchy:

Ratio test. If $L=\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}$ exists, then:

• if $L<1$, the sequence tends to $0$;
• if $L>1$ (also $L=+\infty$), the sequence diverges to $+\infty$;
• if $L=1$, the test says nothing: you need another method.

Root test. If $L=\underset{n\to \infty }{\lim }\sqrt[n]{a_n}$ exists, the same three conclusions as above hold, replacing the ratio with the $n$-th root. The two tests are equivalent in most practical cases; the ratio is almost always more convenient when factorials appear, because $\frac{(n+1)!}{n!}=n+1$ simplifies immediately.

The comparison (squeeze) theorem

When a sequence is “squeezed” between two others that have the same limit, that limit is passed on:

If $b_n\le a_n\le c_n$ eventually (that is, from some $n$ onwards) and ${\lim }_{n\to \infty }{b}_n={\lim }_{n\to \infty }{c}_n=L$, then ${\lim }_{n\to \infty }{a}_n=L$.

In Italy it is nicknamed “the carabinieri theorem” (carabinieri are police officers), and the image is this: $a_n$ is the suspect escorted by two officers who are both going to the same place, so he can do nothing but end up there too. It is the right tool every time a sequence contains a bounded function (like $\sin n$ or $\cos n$) multiplied or divided by something that tends to zero or to infinity: the fourth exercise below is the textbook case.

Four worked exercises

1. A fraction with a root: $\underset{n\to \infty }{\lim }\frac{\sqrt{9n^2+n}}{3n+1}$

I take out $n$ from under the root in the numerator (for $n>0$, $n=\sqrt{n^2}$) and $n$ in the denominator:

As $n\to \infty$, $\frac{1}{n}\to 0$, so the numerator tends to $\sqrt{9}=3$ and the denominator to $3$:

2. Ratio test: $\underset{n\to \infty }{\lim }\frac{2^n}{n!}$

With $a_n=\frac{2^n}{n!}$, I compute the ratio between one term and the previous one:

Since $L=0<1$, by the ratio test $a_n\to 0$. Consistent with the hierarchy of infinities seen above: $n!$ grows faster than any exponential $a^n$, so the ratio between the two is crushed to zero.

3. The standard limit of $e$: $\underset{n\to \infty }{\lim }{\left(1+\frac{3}{n}\right)}^n$

I rewrite the exponent $n$ as $3\cdot \frac{n}{3}$, to bring out the form of the standard limit:

As $n\to \infty$ also $\frac{n}{3}\to \infty$, and ${\left(1+\frac{3}{n}\right)}^{n/3}$ is exactly the standard limit ${\left(1+\frac{1}{m}\right)}^m$ with $m=\frac{n}{3}$, which tends to $e$:

4. Comparison theorem: $\underset{n\to \infty }{\lim }\frac{\sin n}{n}$

The sine is bounded for every value of its argument: $-1\le \sin n\le 1$ for every $n$. Dividing all parts by $n>0$ (the inequality doesn’t change direction, because $n$ is positive):

Both sequences on the sides, $-\frac{1}{n}$ and $\frac{1}{n}$, tend to $0$ as $n\to \infty$. By the comparison theorem, the middle sequence is forced to do the same:

Note that this has nothing to do with ”$\sin n$ has a limit”: $\sin n$ on its own is irregular (it oscillates forever between $-1$ and $1$, like the $(-1{)}^n$ at the start). It is dividing by $n$, which tends to infinity, that squashes the oscillation down to zero.

To practise more exercises of this kind before the midterm, the Calculus 1 practice test covers the whole syllabus with 20 questions: the correct answer to every question is free, the step-by-step guided solution unlocks with the first lesson.

Typical mistakes

• ”$1^{\infty }=1$.” It is an indeterminate form, not a value. The third exercise above is precisely a disguised $1^{\infty }$ ($1+\frac{3}{n}\to 1$, the exponent $\to \infty$) and the true result is $e^3$, not $1$: the base gets close to $1$ but the exponent grows fast enough to change the result. Every time you see a base tending to $1$ raised to an exponent tending to infinity, it is an alarm bell, not a shortcut.
• Confusing a sequence with the function that generates it. $a_n=\sin (\pi n)$ is the constant sequence $0$ (it converges trivially), whereas $f(x)=\sin (\pi x)$ as a function of a real variable has no limit as $x\to \infty$. The step “from function to sequence” is legitimate (if $f(x)\to L$ then $f(n)\to L$), the reverse is not.
• Concluding something from the ratio test (or root test) when $L=1$. In that case the test is silent: you have to change tool, not force a conclusion the test doesn’t give.
• Forgetting “eventually” in the comparison theorem. The inequality $b_n\le a_n\le c_n$ must hold from some index onwards, not necessarily for every $n$ from the start: knowing this avoids wrongly discarding a valid comparison only because the first two or three terms don’t satisfy the inequality.

How I work in lessons

With those who arrive at this chapter, the first job is to separate two things that often stay mixed: recognising what kind of limit you have in front of you (ratio of polynomials, form with factorials, variable exponent, sequence bounded by something that tends to zero) and only then choosing the tool: hierarchy, ratio test, standard limit, comparison. Even before starting the calculations, I always have you compute the first terms by hand: seeing that $\frac{2^n}{n!}$ grows for a while and then collapses teaches more than ten lines of theory on why the factorial wins.

If you want a hand, start from a free 30-minute introductory call: it is there precisely to work out which test is worth concentrating on before the midterm.

For the rest of the Calculus 1 preparation, the natural steps beside sequences are limits of functions, integrals later in the semester and complex numbers, which many courses cover in the same weeks. If the first month of the course has exposed gaps in the high school basics, OFA in maths: how to clear it within the first year deals with the same kind of targeted recovery. The full picture of the subject is on the maths page; university lessons are described at /ripetizioni-universitarie/.