Sequences and Convergence
A sequence is an ordered list of terms \(a_n\). It converges when the terms approach one finite limit \(L\) as \(n \to \infty\) — and limit laws, the Squeeze Theorem, and monotone-bounded convergence are your three tools for pinning \(L\) down.
By the end you'll be able to compute sequence limits with the limit laws, squeeze oscillating sequences like \(\sin(n)/n\) to their limit, and prove a recursively defined sequence converges before you ever compute its limit.
Predict: as \(n\) grows, do the two dashed squeeze bounds \(\pm 1/n\) pinch the dots of \(\sin(n)/n\) toward one value — even though \(\sin(n)\) itself never settles? Slide n to check.
Each dot is one term \(a_n\), plotted against its index \(n\). For \(\sin(n)/n\) the dashed bounds are the squeeze sandwich \(-1/n \le a_n \le 1/n\); the horizontal line is the limit \(L\). Switch to \((3n+1)/(n+2)\) to see a sequence that needs no squeezing — plain limit laws already give its limit. Hover any dot for exact values.
a1 = 0.841 · squeeze bounds: -1.000 ≤ a1 ≤ 1.000 · limit L = 0
A sequence \(\{a_n\}\) converges to \(L\) if its terms get arbitrarily close to \(L\) for large \(n\); limit laws handle algebraic combinations, the Squeeze Theorem tames oscillation, and monotone-bounded convergence proves a limit exists without computing it.
Picture the dots you just explored: a convergent sequence's dots crowd into an ever-thinner horizontal band around \(L\) as you scan right. Divergence looks like dots marching off the top of the chart (\(a_n \to \infty\)) or bouncing forever without settling — \(a_n = (-1)^n\) hops between \(-1\) and \(1\) and never picks a value. The Squeeze Theorem is literally visible: two dashed curves closing like jaws, with the sequence trapped between them. Wherever the jaws meet, the sequence must go.
\(\lim_{n\to\infty} a_n = L\) means \(a_n\) can be made arbitrarily close to \(L\) by taking \(n\) large enough. Limits obey the usual algebra: if \(a_n \to A\) and \(b_n \to B\), then \(a_n \pm b_n \to A \pm B\), \(a_n b_n \to AB\), and \(a_n/b_n \to A/B\) (for \(B \neq 0\)). Since a sequence is a function sampled at integers, you may replace \(n\) with a continuous variable \(x\) and use calculus-1 tools like L'Hôpital's rule — but only after the swap, because L'Hôpital differentiates and you can't differentiate over the integers. Two more workhorses: Squeeze Theorem — if \(a_n \le c_n \le b_n\) eventually and \(\lim a_n = \lim b_n = L\), then \(\lim c_n = L\); and the Monotone Convergence Theorem — every monotone bounded sequence converges. A useful special case: the geometric sequence \(\{r^n\}\) converges exactly when \(-1 < r \le 1\) — to \(0\) if \(|r| < 1\), to \(1\) if \(r = 1\), and diverges otherwise.
Find \(\displaystyle\lim_{n\to\infty} \frac{\sin(n)}{n}\). The factor \(\sin(n)\) oscillates forever, so limit laws stall — but it never leaves \([-1, 1]\). Since \(-1 \le \sin(n) \le 1\), dividing by \(n > 0\) gives $$-\frac{1}{n} \le \frac{\sin(n)}{n} \le \frac{1}{n}.$$ Both outer sequences \(\to 0\), so by the Squeeze Theorem \(\displaystyle\lim_{n\to\infty} \frac{\sin(n)}{n} = \) 0 — exactly the pinch you watched in the interactive above.
Let \(a_1 = \sqrt{2}\) and \(a_{n+1} = \sqrt{2 + a_n}\). There's no closed form for \(a_n\), so first prove a limit exists: the sequence is increasing and bounded above by \(2\), so the Monotone Convergence Theorem guarantees some limit \(L\). Now take the limit of both sides of the recursion: \(L = \sqrt{2 + L}\), so \(L^2 = 2 + L\), giving \(L^2 - L - 2 = 0\). Factor it: \((L - 2)(L + \_\_) = 0\), and since every term is positive, \(L = \_\_\).
Reveal the answer
\(L^2 - L - 2 = (L-2)(L+1) = 0\), so \(L = 2\) or \(L = -1\). The terms are all positive (they start at \(\sqrt2\) and increase), so the limit is \(L = 2\). Notice the two-step structure: the Monotone Convergence Theorem earned you the right to write "\(L\)" at all; only then did algebra find its value.
More info — why prove convergence before computing the limit?
In the recursive example, the step "take the limit of both sides" silently assumed the limit exists. That assumption can lie: apply the same trick to \(a_1 = 1\), \(a_{n+1} = 2a_n\) and you'd "solve" \(L = 2L\) to get \(L = 0\) — but the sequence \(1, 2, 4, 8, \dots\) actually blows up. The Monotone Convergence Theorem is the safety check that makes the algebra legitimate: increasing + bounded above (or decreasing + bounded below) means a limit really is there before you solve for it. Beware the converse trap, too — bounded alone is not enough, as \((-1)^n\) shows. For more practice with this two-step pattern, work the examples in the Paul's Online Math Notes link under Dive deeper.
Check your understanding
Find \(\displaystyle\lim_{n\to\infty} \frac{3n^2+5}{n^2+2}\).
\(a_n = (-1)^n\) is bounded between \(-1\) and \(1\). What can you conclude about its convergence?
Find \(\displaystyle\lim_{n\to\infty} \frac{\cos(n)}{n^2}\).
Find \(\displaystyle\lim_{n\to\infty} \frac{2n + \cos(n)}{n}\). (Think about which tool handles each piece.)
Recap
- A sequence \(\{a_n\}\) converges to \(L\) if \(a_n\) gets arbitrarily close to \(L\) as \(n \to \infty\); otherwise it diverges (blows up or oscillates forever).
- Limit laws mirror function limits: sums, products, and quotients of convergent sequences converge to the sums, products, and quotients of their limits. Swap \(n\) for a continuous \(x\) to borrow L'Hôpital's rule.
- Geometric: \(\{r^n\}\) converges exactly when \(-1 < r \le 1\) — to 0 for \(|r| < 1\), to 1 for \(r = 1\).
- Squeeze Theorem: if \(a_n \le c_n \le b_n\) eventually and both bounds converge to the same \(L\), then \(c_n \to L\) — the tool for oscillating factors like \(\sin(n)\) or \((-1)^n\).
- Monotone Convergence Theorem: monotone + bounded ⇒ convergent — proves a limit exists (essential for recursive sequences) without computing it. Bounded alone is not enough.
Dive deeper
- Paul's Online Math Notes — Sequences Work through convergence, limit laws, and the Squeeze Theorem for sequences.
Sources
- Sequences, Limits, and Convergence