Every iterative method in this course — bisection, Newton's method, Jacobi iteration, the power method — produces a sequence of approximations x_0, x_1, x_2, \ldots. Each one asks the same question. Does it converge, and if so, how fast?
"Converges" just means the terms settle down. They stop moving away from some fixed target. Think of a bouncing ball's peak heights, settling toward zero. This course spends a lot of time on a follow-up question. A first calculus course usually skips it: settling down fast, versus settling down slow. Halving the error every step is very different from barely shrinking it. That difference decides whether an algorithm is worth running, or one that never finishes.
A sequence \{x_n\} converges to a limit L if for every \varepsilon>0 there exists an index N such that |x_n-L|<\varepsilon for every n>N. Eventually the terms stay arbitrarily close to L and never leave. A sequence with no such L diverges. Nothing here is new. What's new later in this course is a sharper question than "does it converge" — how many correct digits does one more iteration buy. That question is about how fast |x_n - L| shrinks, not just whether it does.
A sequence settling toward its limit
Step through the sequence x_n = 1/n term by term and watch how close each new term lands to the limit 0.
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Worked example
Confirming a limit from the definition's spirit
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1
Show that x_n=\frac{n+1}{n} converges to 1, and find how large n must be for |x_n-1|<0.01.
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|x_n - 1| = \left|\frac{n+1}{n}-1\right| = \frac{1}{n}.
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3
Require \frac{1}{n}<0.01, i.e. n>100.
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4\boxed{n \ge 101 \text{ guarantees } |x_n-1|<0.01}
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Worked example
A sequence that oscillates instead of settling
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1
Does x_n=(-1)^n converge?
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The terms are -1,1,-1,1,\ldots — they never settle near a single value, no matter how large n gets.
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3\boxed{\text{diverges (bounded, but never settles)}}
This is the failure mode this course calls oscillation without convergence — an iterative method can produce a bounded sequence of approximations that still never gets closer to an answer.
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Where this goes next
Rate of convergence
A sequence that settles down is the floor, not the ceiling. This course will ask a sharper question, again and again. Does an error sequence e_n = |x_n - L| shrink linearly (e_{n+1}\approx c\,e_n for some c<1)? Does it shrink quadratically (e_{n+1}\approx c\,e_n^2)? Or does it not shrink at all? That's the same idea from this section, now put in numbers. Newton's method owes its whole fame for speed to landing in that faster case.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
Find \displaystyle\lim_{n\to\infty} \dfrac{7n+3}{n+9}.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Divide top and bottom by n: (7+3/n)/(1+9/n) \to (7+0)/(1+0) = 7.
Correct answer
Also accepted:
Why
Divide numerator and denominator by the highest power of n present.
From Calculus II — Integral Calculus & Infinite Series
Find this same-degree rational-sequence limit: \displaystyle\lim_{n\to\infty} \dfrac{4n^2-1}{2n^2+n}.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
Divide by n^2: (4-1/n^2)/(2+1/n) \to 4/2 = 2.
Correct answer
Also accepted:
Why
Divide numerator and denominator by n squared.
From Calculus II — Integral Calculus & Infinite Series
A sequence whose terms grow without bound as n\to\infty is said to...
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
"Converge" is reserved for settling at a finite number -- infinity is not a finite limit.
Oscillation means bouncing between values, not racing off in one direction.
An unbounded sequence cannot be bounded.
Why
A sequence with no finite limit diverges; "diverges to infinity" specifically describes the unbounded-growth case, as opposed to oscillating divergence.
Convergence always means settling at a finite number.