Sequences and Convergence

In this lesson you’ll learn the epsilon-N definition of a limit, the standard template for a convergence proof, why limits are unique, and that convergent sequences are bounded. This is the first rigorous definition in the section and the pattern for all the others.

A sequence $(a_n)$ converges to $L \in \mathbb{R}$ if

$$\forall \varepsilon > 0 \; \exists N \in \mathbb{N} \; \text{such that} \; \forall n > N, \; |a_n - L| < \varepsilon$$

Read it as a game. Your opponent picks $\varepsilon$, as small as they like. You must produce an $N$ such that from that point on, every term is within $\varepsilon$ of $L$. If you can always win, the sequence converges.

The picture is the definition: draw a horizontal band of half-width $\varepsilon$ around $L$. Convergence means that however thin the band, only finitely many terms fall outside it, and every term past $N$ is inside.

Three things to notice.

• $N$ depends on $\varepsilon$. Smaller $\varepsilon$ means larger $N$. That dependence is the whole content.
• $N$ need not be optimal. Any $N$ that works is fine. Proofs routinely use a wasteful $N$ because it’s easier to write down.
• Finitely many terms may misbehave. The definition only constrains what happens eventually, so changing the first million terms changes nothing.

Every convergence proof has the same three parts.

1. Scratch work (not written). Start from $|a_n - L| < \varepsilon$ and solve for $n$ to see how large $n$ must be.
2. Choose $N$. State it, usually via the Archimedean property.
3. Verify. Assume $n > N$ and derive $|a_n - L| < \varepsilon$ in the forward direction.

Take $a_n = \frac{1}{n} \to 0$.

Scratch. $\left|\frac1n - 0\right| = \frac1n < \varepsilon \iff n > \frac1\varepsilon$.

Proof. Let $\varepsilon > 0$. By the Archimedean property choose $N \in \mathbb{N}$ with $N > \frac1\varepsilon$. Then for $n > N$,

$$\left|\frac1n - 0\right| = \frac1n < \frac1N < \varepsilon$$

Hence $\frac1n \to 0$. ∎

Short, and the shape never changes. The scratch work is where the thinking happens; the written proof is the verification only.

$(a_n)$ diverges if it converges to no real number. Negating the definition:

$$\forall L \; \exists \varepsilon>0 \; \text{such that} \; \forall N \; \exists n > N \text{ with } |a_n - L| \ge \varepsilon$$

So to prove divergence you must, for every candidate $L$, exhibit an $\varepsilon$ that is violated infinitely often.

Three flavours:

• To infinity. $a_n = n$. For every $M$ there is $N$ with $a_n > M$ for $n > N$. Written $a_n \to \infty$, which is a description of divergence, not a limit.
• By oscillation. $a_n = (-1)^n$. Bounded, never settles.
• Unbounded and oscillating. $a_n = (-1)^n n$.

If $a_n \to L$ and $a_n \to M$ then $L = M$.

Proof. Suppose $L \ne M$ and set $\varepsilon = \frac{|L - M|}{2} > 0$.

There is $N_1$ with $|a_n - L| < \varepsilon$ for $n > N_1$, and $N_2$ with $|a_n - M| < \varepsilon$ for $n > N_2$. Take $n > \max(N_1, N_2)$. By the triangle inequality,

$$|L - M| \le |L - a_n| + |a_n - M| < \varepsilon + \varepsilon = |L - M|$$

so $|L-M| < |L-M|$, a contradiction. Hence $L = M$. ∎

Two techniques there are worth naming, because they recur constantly: taking the max of two thresholds so both conditions hold at once, and inserting a term via the triangle inequality to bridge two quantities.

Every convergent sequence is bounded.

Proof. Say $a_n \to L$. Apply the definition with $\varepsilon = 1$: there is $N$ with $|a_n - L| < 1$ for $n > N$, so $|a_n| < |L| + 1$ for those $n$.

The remaining terms $a_1, \dots, a_N$ are a finite list, so they have a largest absolute value. Then

$$M = \max\{|a_1|, \dots, |a_N|, |L| + 1\}$$

bounds the whole sequence. ∎

The converse fails: $(-1)^n$ is bounded and divergent. Repairing that gap is Bolzano-Weierstrass, two lessons ahead.

Note the structure: handle the tail with the definition, the head by finiteness. That split is a standard move.

A subsequence picks out infinitely many terms in order: $a_{n_1}, a_{n_2}, \dots$ with $n_1 < n_2 < \cdots$.

If $a_n \to L$ then every subsequence also $\to L$. Contrapositive: two subsequences with different limits proves divergence. For $(-1)^n$, the even terms give 1 and the odd terms give $-1$, so it diverges - much quicker than negating the definition directly.

Example 1: Prove $\frac{2n+1}{n} \to 2$.

Solution. Scratch. $\left|\frac{2n+1}{n} - 2\right| = \left|\frac{2n+1-2n}{n}\right| = \frac1n$, and $\frac1n < \varepsilon \iff n > \frac1\varepsilon$.

Proof. Let $\varepsilon>0$ and choose $N > \frac1\varepsilon$. For $n > N$,

$$\left|\frac{2n+1}{n} - 2\right| = \frac1n < \frac1N < \varepsilon$$

∎

Example 2: Prove $\frac{n}{n+1} \to 1$.

Solution. Scratch. $\left|\frac{n}{n+1} - 1\right| = \left|\frac{n - n - 1}{n+1}\right| = \frac{1}{n+1}$, which is $< \varepsilon$ when $n > \frac1\varepsilon - 1$.

Proof. Given $\varepsilon>0$, choose $N > \frac1\varepsilon$. For $n > N$,

$$\left|\frac{n}{n+1} - 1\right| = \frac{1}{n+1} < \frac1n < \frac1N < \varepsilon$$

∎

The step $\frac{1}{n+1} < \frac1n$ is a deliberate over-estimate. Simplifying in the direction that makes the bound larger is always legal and usually saves work.

Example 3: A sequence needing a sharper estimate.

Prove $\frac{3n^2}{n^2+5} \to 3$.

Solution. Scratch.

$$\left|\frac{3n^2}{n^2+5} - 3\right| = \left|\frac{3n^2 - 3n^2 - 15}{n^2+5}\right| = \frac{15}{n^2+5}$$

Rather than solve exactly, over-estimate: $\frac{15}{n^2+5} < \frac{15}{n^2} \le \frac{15}{n}$, and $\frac{15}{n} < \varepsilon \iff n > \frac{15}{\varepsilon}$.

Proof. Let $\varepsilon>0$, choose $N > \frac{15}{\varepsilon}$. For $n > N$,

$$\left|\frac{3n^2}{n^2+5} - 3\right| = \frac{15}{n^2+5} < \frac{15}{n} < \frac{15}{N} < \varepsilon$$

∎

Example 4: Prove $(-1)^n$ diverges, twice.

Solution 1 (subsequences). The even-indexed subsequence is constantly $1$, so it converges to 1. The odd-indexed subsequence is constantly $-1$, converging to $-1$. Different limits, so the sequence diverges. ∎

Solution 2 (definition). Let $L$ be any real and take $\varepsilon = 1$. Consecutive terms differ by 2, so for any $n$,

$$|a_n - L| + |a_{n+1} - L| \ge |a_n - a_{n+1}| = 2$$

meaning at least one of the two is $\ge 1$. So beyond any $N$ there is a term with $|a_n - L| \ge 1$, and no $L$ works. ∎

The subsequence argument is shorter and it’s the one to reach for.

Example 5: A convergent sequence with a badly behaved head.

Let $a_n = 10^6$ for $n \le 500$ and $a_n = \frac1n$ for $n > 500$. Does it converge?

Solution. Yes, to 0. Given $\varepsilon>0$, choose $N > \max\left(500, \frac1\varepsilon\right)$. For $n > N$ we have $n > 500$, so $a_n = \frac1n < \varepsilon$. ∎

Convergence is entirely a statement about the tail. Any finite number of terms can be arbitrary.

Example 6: Reading off the wrong quantifier order.

What would "$\exists N \; \forall \varepsilon>0 \; \forall n>N: |a_n - L|<\varepsilon$" describe?

Solution. A single $N$ working for every $\varepsilon$ at once forces $|a_n - L| < \varepsilon$ for all $\varepsilon>0$, hence $|a_n - L| = 0$, so $a_n = L$ exactly for all $n > N$.

That describes a sequence eventually constant at $L$ - far stronger than convergence, and it would exclude $\frac1n \to 0$. The quantifier order is the definition’s whole content.

Iterative algorithms. Newton’s method, gradient descent, and the power method for eigenvalues all produce sequences, and the question “does it converge, and how fast?” is exactly the epsilon-N question. A stopping criterion like “halt when successive iterates differ by less than $10^{-8}$” is choosing an $\varepsilon$ and hoping for the corresponding $N$.

Fixed-point iteration in engineering. Solving a circuit or a fluid model iteratively converges only under conditions like a contraction property, which is proved with these tools. When a simulation oscillates instead of settling, the sequence is diverging by oscillation.

Machine learning training curves. A loss curve is a sequence. Distinguishing “converged,” “converging slowly,” and “oscillating” is what learning-rate schedules are managing, and the reason a too-large learning rate makes loss oscillate is a genuine divergence-by-oscillation.

Financial models. Present value of a perpetuity is the limit of a sequence of partial sums, and whether it converges determines whether the asset has finite value. Discount rates below the growth rate make the sequence diverge, which is how bubble models get their infinite valuations.

Monte Carlo simulation. The running average of samples is a sequence converging to the true expectation. The rate, $O(1/\sqrt{n})$, tells you that reducing error tenfold takes a hundred times more samples, which is a practical consequence of a convergence rate.