Calculus I · Patterns That Go Somewhere

Three fates

Not every sequence converges. There are exactly three things a sequence can do, and naming them honestly keeps you from forcing a limit where there is none.

It can converge — settle to a single number $L$, like $\tfrac1n \to 0$.

It can diverge to infinity — grow without bound, like $a_n = n^2$ or like a geometric sequence with $|r| > 1$. There is no limit; the terms outrun every ceiling.

It can oscillate — bounce around and never settle, like $a_n = (-1)^n$, which is $-1, 1, -1, 1, \ldots$ forever. It stays bounded but it has no limit, because it never gets and stays close to any one number. (It is near $1$ half the time and near $-1$ the other half; “near $L$ half the time” is not convergence.)

That last case is the warning the rest of this section answers. Squinting at a table can fool you — a few terms can look like they’re settling when they’re not. To be sure, we need to say exactly what “gets and stays arbitrarily close” means. That is Layer 2.

◆ Layer 2 — What “approaches” really means

Here is the definition that makes calculus rigorous. It looks intimidating for about a day and then becomes the most natural thing in the world. Read it as a game between two players.

Definition. $a_n \to L$ means: for every tolerance $\varepsilon > 0$, there exists a cutoff $N$ such that for all $n > N$, $\;|a_n - L| < \varepsilon$.

The game. A skeptic hands you a tolerance $\varepsilon$ — say, “I bet your terms don’t get within $\varepsilon = 0.001$ of $L$ and stay there.” Your job is to answer with a cutoff $N$ — a seat number — and promise: “every term from seat $N$ onward is within $0.001$ of $L$.” If you can win this game no matter how tiny a tolerance the skeptic names, then $a_n \to L$ is true. The quantity $|a_n - L|$ is just the distance from the term to the target; $|a_n - L| < \varepsilon$ says “this term is inside a window of width $\varepsilon$ around $L$.”

Why this and not “gets closer and closer”? Because “gets closer and closer” is a trap: the sequence $1, 0.9, 0.99, 1.001, 0.5, \ldots$ might get closer for a while and then wander off, and the vague phrase can’t tell. The $\varepsilon$–$N$ definition forbids wandering off: past the cutoff, every term must stay in the window. That single word “every” is what your intuition was missing.

◆ Proof: $\dfrac1n \to 0$

We play the game. Let a tolerance $\varepsilon > 0$ be handed to us. We must produce a cutoff $N$ so that for all $n > N$, the distance $\left|\tfrac1n - 0\right| = \tfrac1n$ is less than $\varepsilon$.

We want $\dfrac1n < \varepsilon$. Rearranging, that is the same as $n > \dfrac1\varepsilon$. So we simply choose $N$ to be any whole number with $N \ge \dfrac1\varepsilon$. (Such a whole number always exists — the number line never runs out of integers above a given value. That’s a basic property of the reals, sometimes called the Archimedean property.)

Now check it works: if $n > N \ge \dfrac1\varepsilon$, then $n > \dfrac1\varepsilon$, which rearranges back to $\dfrac1n < \varepsilon$. So every term past seat $N$ is within $\varepsilon$ of $0$. Since we won the game for an arbitrary $\varepsilon$, the limit is $0$. $\blacksquare$

Sit with this one. It is the first real $\varepsilon$–$N$ proof in the book, and almost every other convergence proof is this same three-move dance: let $\varepsilon$ be given → solve the inequality for $n$ → choose $N$ accordingly and verify.

◆ Proof: $r^n \to 0$ when $|r| < 1$

This is the fact the whole of Chapter 8 will lean on, so it deserves a real proof. Assume $0 < |r| < 1$ (the case $r = 0$ is trivially $0$).

Since $|r| < 1$, its reciprocal $\tfrac{1}{|r|}$ is bigger than $1$, so we can write $\tfrac{1}{|r|} = 1 + h$ for some $h > 0$. Then $|r|^n = \dfrac{1}{(1+h)^n}$. Now use Bernoulli’s inequality, $(1+h)^n \ge 1 + nh$ (proved in problem 30), to push the denominator up and the whole fraction down:

$$|r|^n = \frac{1}{(1+h)^n} \le \frac{1}{1 + nh} < \frac{1}{nh}.$$

Now play the game. Given $\varepsilon > 0$, we want $\tfrac{1}{nh} < \varepsilon$, i.e. $n > \tfrac{1}{h\varepsilon}$. Choose $N \ge \tfrac{1}{h\varepsilon}$. Then for every $n > N$ we have $|r^n| = |r|^n < \tfrac{1}{nh} < \varepsilon$. So $r^n \to 0$. $\blacksquare$

The engine is the same as before — the only cleverness was rewriting $|r|$ as $\tfrac{1}{1+h}$ so that Bernoulli’s inequality could turn the shrinking power into an honest $\tfrac1n$-style bound we already know how to beat.

◆ Layer 2 — Sequences that must arrive (and the birth of $e$)

Sometimes you can see a sequence climbing steadily but you cannot compute its limit directly — exactly the situation with $\left(1+\tfrac1n\right)^n$. There is a beautiful principle that rescues us, and it comes straight from the property of the real number line you met in Book 1: the line has no gaps.

The Monotone Convergence principle. A sequence that is increasing and bounded above (never climbs past some ceiling) must converge to a limit. (Likewise: decreasing and bounded below must converge.)

The picture is honest and the gaplessness is doing the work: an increasing sequence trapped under a ceiling has nowhere to escape to, and because the real line has no holes for it to pile up against, there is a genuine number it closes in on. (On a number line with gaps — like the rationals alone — this can fail, which is precisely the defect Book 1’s irrational numbers fixed.)

Now point it at our mystery sequence. The sequence $\left(1+\tfrac1n\right)^n$ turns out to be increasing (each term beats the last — take this on faith here; it’s true and provable) and bounded above by $3$ (it never reaches $3$ — you’ll prove this yourself in problem 31, and the proof reuses the geometric sequence $1 + \tfrac12 + \tfrac14 + \cdots$ from this very chapter). Increasing and bounded above: by the principle, it must converge. Its limit is a real, specific number, and since it doesn’t equal anything we’ve already named, we give it a new name:

$$e = \lim_{n\to\infty}\left(1 + \frac1n\right)^n \approx 2.71828\ldots$$

Notice what just happened: we proved $e$ exists without ever computing it exactly — the principle guarantees a limit is there to be found. That is a new kind of power. We will discover in Chapter 5 that this same $e$ is the one base whose exponential curve is its own slope, and we’ll prove the two descriptions match in Chapter 10. Born here, fully understood three chapters from now.

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