Convergence of sequences
A list of numbers can creep toward a value, swing back and forth forever, or run off without bound. What does it mean, exactly, for the terms to settle on one number, and how can we prove that they settle when there is no formula to compute with?
Where you are 8 in Sequences and series in calculus
Why this exists
A sequence (Sequences, terms and notation) is an endless list a_1, a_2, a_3, \ldots, given by a formula or by a rule that builds each term from the one before (Generating a sequence from a rule). Calculus asks about the far end of the list: do the terms settle on one number?
Levani takes 50 milligrams of a medicine at equal intervals, and by the next dose his body has cleared 40\% of what was there. The amounts just after each dose are 50, 80, 98, 108.8, \ldots, each found from the last by a_{n+1} = 0.6a_n + 50. Does the amount level off, and at what? There is no formula for a_n to plug into.
Day 1 makes “settles” exact and gives the laws that compute limits of sequences given by formulas, much as limits at infinity did for functions. Day 2 proves that a sequence converges without first knowing its limit: a sequence that only climbs, and never passes some ceiling, must settle. That fact rests on a property of the real numbers themselves. Both tools are needed next, where the sequence being tracked is the running total of an infinite sum.
Definition
In plain words: a sequence converges to L if its terms eventually stay as close to L as anyone demands. Not close once, but close from some point on, forever.
In symbols: the sequence \{a_n\} converges to the number L, written
if for every number \varepsilon > 0 there is a number N such that
The number L is the limit. A sequence that converges to no number diverges. If for every M the terms satisfy a_n > M from some point on, the sequence diverges to infinity, written \lim_{n \to \infty} a_n = \infty; the limit does not exist as a number, and the notation says how it fails.
The tolerance \varepsilon is named first; the cutoff N is then found to answer it, and may depend on \varepsilon. This is the definition of \lim_{x \to \infty} f(x) with the input restricted to the positive integers.
Monotone and bounded (Day 2). A sequence is nondecreasing if a_n \le a_{n+1} for every n, and increasing if a_n < a_{n+1} for every n; nonincreasing and decreasing reverse the inequalities. A sequence that is nondecreasing or nonincreasing is monotone. It is bounded above if some number M has a_n \le M for every n, bounded below if some m has a_n \ge m for every n, and bounded if both. A sequence is recursively defined when a first term and a rule a_{n+1} = f(a_n) are given.
Intuition
Draw the sequence as dots (n, a_n) (Figure 1). Shade a band of half-width \varepsilon around the line y = L. Convergence says: however thin the band, from some N on every dot lies inside it. For \displaystyle a_n = \frac{3n^2 - n}{5n^2 + 2} and \varepsilon = 0.05 the dots enter the band at n = 5 and never leave. Only the existence of some N matters, not the smallest one.
A squeeze looks as it did for functions in The squeeze theorem and the limit of sine x over x: the dots of \displaystyle \frac{\cos n}{n} bounce up and down, but they are trapped between \displaystyle -\frac{1}{n} and \displaystyle \frac{1}{n}, which close in on 0 (Figure 2).
The amounts after each dose (Figure 3) climb, each climb smaller than the last, and never pass 125 milligrams. A climbing sequence with a ceiling has nowhere to go but toward some number at or below the ceiling, because the real number line has no gaps for it to fall into. That is the completeness of the real numbers, the one property Day 2 takes on trust.
Results
Day 1: limits of sequences given by formulas
Uniqueness. A sequence has at most one limit.
Limit laws. If a_n \to A and b_n \to B, then a_n + b_n \to A + B, a_n - b_n \to A - B, c\,a_n \to cA for any constant c, a_n b_n \to AB, and \displaystyle \frac{a_n}{b_n} \to \frac{A}{B} provided B \ne 0. If g is continuous at A, then g(a_n) \to g(A); this covers powers, roots where defined, e^{a_n}, and \ln a_n when A > 0.
From functions to sequences. If \lim_{x \to \infty} f(x) = L and a_n = f(n) for every n, then \lim_{n \to \infty} a_n = L. The same holds with L = \infty.
Squeeze theorem for sequences. If a_n \le b_n \le c_n for every n beyond some N_0, and a_n \to L and c_n \to L, then b_n \to L. In particular, if |b_n| \to 0, then b_n \to 0.
Geometric sequences. r^n \to 0 if |r| < 1, r^n \to 1 if r = 1, and \{r^n\} diverges for every other r.
Finitely many terms do not matter. Changing, adding or dropping finitely many terms changes neither whether a sequence converges nor its limit. In particular, if a_n \to L then a_{n+1} \to L.
Day 2: monotone sequences
Convergent sequences are bounded.
Completeness of the real numbers (stated without proof). Every nonempty set of real numbers that is bounded above has a least upper bound: an upper bound smaller than every other upper bound. This is part of how the real numbers are built, and it is assumed here, not proved.
Monotone convergence theorem. A nondecreasing sequence that is bounded above converges, and its limit is the least upper bound of its terms. A nonincreasing sequence that is bounded below converges to the greatest lower bound of its terms. Since finitely many terms do not matter, it is enough that the sequence be monotone from some term on.
Limits of recursive sequences. If a_{n+1} = f(a_n), the sequence converges to L, and f is continuous at L, then L = f(L).
Why it is true
Sum law. Let \varepsilon > 0. There is N_1 with \displaystyle |a_n - A| < \frac{\varepsilon}{2} for n > N_1, and N_2 with \displaystyle |b_n - B| < \frac{\varepsilon}{2} for n > N_2. For n beyond both, the triangle inequality gives
The other laws and uniqueness are proved as for functions in The epsilon delta definition of a limit, with “for n > N” in place of “for 0 < |x - a| < \delta”.
From functions to sequences. Given \varepsilon > 0, the limit of f supplies a number M with |f(x) - L| < \varepsilon for every real x > M. The integers n > M are among those x, so N = M works.
Squeeze. Given \varepsilon > 0, beyond some cutoff all three hold: a_n \le b_n \le c_n, a_n > L - \varepsilon, and c_n < L + \varepsilon. There L - \varepsilon < b_n < L + \varepsilon. For the special case, use -|b_n| \le b_n \le |b_n|.
Geometric sequences. For 0 < r < 1, r^x = e^{x \ln r} with \ln r < 0, so r^x \to 0 as x \to \infty, and the function link gives r^n \to 0. For r = 0 every term is 0. For -1 < r < 0, |r^n| = |r|^n \to 0, and the special case of the squeeze applies. For |r| > 1 the terms are unbounded, so they cannot converge (Day 2). For r = -1 the terms alternate between -1 and 1, and no number is within 1 of both, so taking \varepsilon = 1 defeats every proposed limit.
Convergent means bounded. Take \varepsilon = 1: beyond some N, every term lies between L - 1 and L + 1. Only finitely many terms come before, so the largest and smallest of those terms, together with L \pm 1, bound the whole sequence.
Monotone convergence. Let \{a_n\} be nondecreasing and bounded above. By completeness its terms have a least upper bound L. Let \varepsilon > 0. The number L - \varepsilon is less than L, so it is not an upper bound: some term has a_N > L - \varepsilon. For every n > N, a_N \le a_n \le L, because the sequence never decreases and L is an upper bound. So |a_n - L| < \varepsilon. For a nonincreasing sequence bounded below, apply this to \{-a_n\}.
Recursive limits. If a_n \to L, then a_{n+1} \to L, since it is the same list with one term dropped, and f(a_n) \to f(L) by continuity. But a_{n+1} and f(a_n) are the same sequence, and a limit is unique, so L = f(L).
Worked examples
Day 1: Eka’s rational sequence, from the laws and from the definition
Eka wants \lim_{n \to \infty} \dfrac{3n^2 - n}{5n^2 + 2}, a proof from the definition, and an N that works for \varepsilon = 0.01.
SolutionDivide top and bottom by n^2, the highest power in the bottom:
Since \displaystyle \frac{1}{n} \to 0 and \displaystyle \frac{2}{n^2} \to 0, the difference, sum and quotient laws (the bottom limit 5 is not 0) give \displaystyle a_n \to \frac{3}{5}.
For the proof, put the distance to \displaystyle \frac{3}{5} over a common denominator:
It is enough to bound this by something simpler. For n \ge 2, 6 \le 3n, so the top in size is at most 8n, and the bottom is more than 25n^2. Hence
and \displaystyle \frac{8}{25n} < \varepsilon once \displaystyle n > \frac{8}{25\varepsilon}. So, given \varepsilon > 0, let N be the larger of 2 and \displaystyle \frac{8}{25\varepsilon}; every n > N has \displaystyle \left|a_n - \frac{3}{5}\right| < \varepsilon.
For \varepsilon = 0.01 this gives N = 32. It is not the smallest cutoff: solving \displaystyle \frac{5n + 6}{25n^2 + 10} < 0.01 exactly shows that every term from n = 22 on is already within 0.01. The definition asks only for some N.
Day 1: a difference of two large quantities
Find \displaystyle \lim_{n \to \infty} \left(\sqrt{n^2 + n} - n\right).
SolutionBoth \displaystyle \sqrt{n^2 + n} and n grow without bound, and the difference law needs two limits that are numbers, so it does not apply. Multiply and divide by the conjugate \displaystyle \sqrt{n^2 + n} + n:
Divide top and bottom by n, using \displaystyle \frac{\sqrt{n^2 + n}}{n} = \sqrt{1 + \frac{1}{n}} for n > 0:
The square root is continuous at 1, so \displaystyle \sqrt{1 + \frac{1}{n}} \to 1, and the quotient law gives the limit \displaystyle \frac{1}{2}. At n = 100 the term is about 0.4988. The two quantities grow side by side, and the gap between them settles at one half.
Day 1: Dato’s power that pulls two ways
Dato wants \lim_{n \to \infty} \left(1 - \dfrac{1}{n}\right)^n.
SolutionThe base tends to 1, which pulls the power toward 1, while the growing exponent pulls it toward 0. No law covers a variable exponent, so pass to the function \displaystyle f(x) = \left(1 - \frac{1}{x}\right)^x for x > 1 and take its logarithm, with \displaystyle t = \frac{1}{x}:
As x \to \infty, t \to 0^+. The quotient \displaystyle \frac{\ln(1 - t) - \ln 1}{t - 0} is a difference quotient of g(t) = \ln(1 - t) at 0, so it tends to \displaystyle g'(0) = -\frac{1}{1 - 0} = -1 (L’Hopital’s rule agrees). Since the exponential is continuous, f(x) = e^{\ln f(x)} \to e^{-1}, and the function link gives
The approach is slow: (0.9)^{10} \approx 0.3487 and (0.99)^{100} \approx 0.3660. Compare \displaystyle \left(1 + \frac{1}{n}\right)^n \to e in The number e and the natural exponential function.
Day 1: a wobble divided by n
Find \lim_{n \to \infty} \dfrac{\cos n}{n}, then \lim_{n \to \infty} \dfrac{5n + \cos n}{2n + \cos n}.
SolutionThe product law would need a limit for \cos n, and nothing so far says it has one (it has none). Squeeze instead. For every n, -1 \le \cos n \le 1; dividing by the positive number n keeps the inequalities:
Both walls tend to 0, so \displaystyle \frac{\cos n}{n} \to 0 (Figure 2).
For the second sequence, the bottom is at least 2n - 1 > 0, so every term is defined. Divide top and bottom by n:
The top tends to 5 and the bottom to 2 \ne 0, so by the quotient law the limit is \displaystyle \frac{5}{2}.
Day 2: factorials against powers
Show that a_n = \dfrac{n!}{n^n} converges by the monotone convergence theorem, then find its limit.
SolutionThe first terms are \displaystyle 1, \frac{1}{2}, \frac{2}{9}, \frac{3}{32}. Compare consecutive terms through their ratio, using (n + 1)! = (n + 1)\,n!:
Since \displaystyle \frac{n}{n+1} < 1, the ratio is less than 1, and the terms are positive, so a_{n+1} < a_n: the sequence is decreasing and bounded below by 0. The monotone convergence theorem says it converges to some L \ge 0.
The theorem gives existence, not the value. One way to find it: a_{n+1} = r_n a_n with \displaystyle r_n = \left(\frac{n}{n+1}\right)^n = \frac{(1 - \frac{1}{n+1})^{n+1}}{1 - \frac{1}{n+1}}, which tends to \displaystyle \frac{1/e}{1} = \frac{1}{e} by Example 3. Taking limits, \displaystyle L = \frac{L}{e}, so L = 0. A squeeze confirms it directly, since every factor after the first is at most 1:
Day 2: Levani’s medication levels off
With a_1 = 50 and a_{n+1} = 0.6a_n + 50, show that the amount just after each dose converges, find the limit, and find the first dose after which the amount is within 1 milligram of it.
SolutionIf the amount levels off at L, then L = 0.6L + 50, so L = 125. That is only a guess until convergence is proved.
Bounded above by 125. First, a_1 = 50 < 125. If a_n < 125, then a_{n+1} = 0.6a_n + 50 < 0.6 \cdot 125 + 50 = 125. By induction every term is below 125.
Increasing. a_{n+1} - a_n = 50 - 0.4a_n, which is positive because a_n < 125.
By the monotone convergence theorem the sequence converges to some L. The function f(x) = 0.6x + 50 is continuous, so L = 0.6L + 50 and L = 125 milligrams. That is the level at which the 40\% cleared between doses, 50 milligrams, exactly replaces the dose.
For the cutoff, subtract 125 = 0.6 \cdot 125 + 50 from the rule: a_{n+1} - 125 = 0.6(a_n - 125). The gap shrinks by the factor 0.6 at each dose, starting from a_1 - 125 = -75:
The amount is within 1 milligram when \displaystyle (0.6)^{n-1} < \frac{1}{75}, that is \displaystyle n - 1 > \frac{\ln 75}{\ln (5/3)} \approx 8.45. So the first such dose is n = 10: a_9 \approx 123.74 and a_{10} \approx 124.24.
Day 2: a nested square root
Let a_1 = 1 and \displaystyle a_{n+1} = \sqrt{2 + a_n}. Show that the sequence converges and find its limit.
SolutionThe terms \displaystyle a_2 = \sqrt{3} \approx 1.7321, a_3 \approx 1.9319, a_4 \approx 1.9829 suggest a climb toward 2.
Bounded. Every term after the first is a square root, and a_1 = 1, so all terms are positive. If 0 < a_n < 2, then \displaystyle a_{n+1} < \sqrt{4} = 2. By induction 0 < a_n < 2 for every n.
Increasing. Both terms are positive, so a_{n+1} > a_n exactly when a_{n+1}^2 > a_n^2, that is 2 + a_n > a_n^2, or (a_n - 2)(a_n + 1) < 0. This holds because 0 < a_n < 2.
The sequence converges to some L, and L \ge a_1 = 1 since it is increasing. The function \displaystyle \sqrt{2 + x} is continuous for x \ge -2, so \displaystyle L = \sqrt{2 + L}, giving L^2 - L - 2 = 0 and L = 2 or L = -1. The second is impossible, so the limit is 2.
A closed form agrees. Since \displaystyle 2 + 2\cos\theta = 4\cos^2 \frac{\theta}{2}, for 0 \le \theta \le \pi we have \displaystyle \sqrt{2 + 2\cos\theta} = 2\cos\frac{\theta}{2}. Starting from \displaystyle a_1 = 2\cos\frac{\pi}{3}, each step halves the angle, so \displaystyle a_n = 2\cos \frac{\pi}{3 \cdot 2^{n-1}} \to 2\cos 0 = 2.
Boundaries
- From functions to sequences runs one way. \sin(\pi n) = 0 for every n, so that sequence converges to 0, yet \sin(\pi x) has no limit as x \to \infty. A sequence can converge when the function through it does not.
- The quotient law needs B \ne 0, and the difference law needs both limits to be numbers (Example 2).
- Bounded alone, or monotone alone, is not enough. (-1)^n is bounded and diverges; a_n = n is increasing and diverges. The monotone convergence theorem needs both.
- Solve L = f(L) only after convergence is proved. With a_1 = 1 and a_{n+1} = 2a_n + 1, the equation gives L = -1, but the terms 1, 3, 7, 15, \ldots grow without bound. And when L = f(L) has several roots, bounds on the terms must pick one (Example 7).
- The squeeze walls must share a limit. -1 \le (-1)^n \le 1 traps nothing.
- Completeness is essential. Among rational numbers alone, 1, 1.4, 1.41, 1.414, \ldots (the decimal truncations of \displaystyle \sqrt{2}) is increasing and bounded above, but has no rational limit.
Where this goes next
The sum of an infinite series is defined as the limit of the sequence of its partial sums, so every question about series is a question about a sequence. When the terms added are positive, the partial sums are nondecreasing, and the monotone convergence theorem is what makes the Convergence tests work: bound the partial sums and convergence follows. The geometric sequence r^n is behind the The geometric series test, and the limit of the terms themselves drives the The nth term test for divergence. Linear recursions like the dosing rule are solved in general in Solving and checking a two-term recurrence; the same dosing total, written as a sum, appears in Infinite geometric series.
Exercises
Day 1: limits from the laws
Find each limit, or say how the sequence diverges.
- \lim_{n \to \infty} \dfrac{2n^3 + 5}{4n^3 - n^2}
- \lim_{n \to \infty} \dfrac{7 - 4n}{3n + 1}
- \lim_{n \to \infty} \dfrac{n^2 + 1}{2n + 3}
- \displaystyle \lim_{n \to \infty} \left(\sqrt{4n^2 + 3n} - 2n\right)
- \lim_{n \to \infty} \dfrac{3^n + 2}{3^{n+1} - 1}
- \lim_{n \to \infty} \left(4 + \left(-\dfrac{2}{3}\right)^n\right)
- \lim_{n \to \infty} \dfrac{(-1)^n n}{n + 1}
Day 1: limits through functions
- \lim_{n \to \infty} \big(\ln(3n + 1) - \ln n\big)
- \lim_{n \to \infty} \left(1 - \dfrac{2}{n}\right)^{3n}
- \lim_{n \to \infty} n \sin \dfrac{1}{n}
- \lim_{n \to \infty} n^{1/n}
Day 1: squeezes and the definition
- \displaystyle \lim_{n \to \infty} \dfrac{\sin(n^2)}{\sqrt{n}}
- \lim_{n \to \infty} \dfrac{4n - \sin(3n)}{n + 2}
- \lim_{n \to \infty} \dfrac{2^n}{n!}
- Prove from the definition that \dfrac{2n + 1}{n + 3} \to 2, and find the smallest integer N such that every n > N has |a_n - 2| < 0.001.
Day 2: monotone and bounded
- Show that a_n = \dfrac{3n}{n + 4} is increasing and bounded, and find its limit.
- Show that a_n = \dfrac{n}{2^n} is nonincreasing and bounded below, and find its limit.
- Describe where a_n = \dfrac{5^n}{n!} increases and decreases, explain why it converges, and find its limit.
Day 2: recursive sequences
- Eka takes 120 milligrams of a medicine at equal intervals, and 70\% of the amount in her body remains at the next dose: a_1 = 120 and a_{n+1} = 0.7a_n + 120. Show the sequence is increasing and bounded above by 400, find its limit, and find the first dose after which the amount exceeds 390 milligrams.
- Dato’s treatment starts with a loading dose of 300 milligrams, then 80 milligrams at each later dose, with half the amount cleared between doses: a_1 = 300 and a_{n+1} = 0.5a_n + 80. Show the sequence is decreasing and bounded below, and find its limit.
- Let a_1 = 1 and \displaystyle a_{n+1} = \sqrt{3a_n}. Show the sequence converges and find its limit.
- Levani studies a_1 = 1 and a_{n+1} = \dfrac{1}{2}\left(a_n + \dfrac{3}{a_n}\right). Show that from a_2 on the sequence is nonincreasing and bounded below by \displaystyle \sqrt{3}, and find its limit.
Reasoning
- Dato claims that every bounded sequence converges, and Eka claims that every monotone sequence converges. Give a counterexample to each claim, and state the theorem that the two hypotheses make true together.
- Prove that if a_n \to L then |a_n| \to |L|. Show by example that \{|a_n|\} can converge while \{a_n\} does not, and explain why the converse does hold when L = 0.
- The sequence a_1 = 2, a_{n+1} = a_n^2 has the fixed-point equation L = L^2, with roots 0 and 1. Does the sequence converge to either? Explain what the fixed-point equation does and does not tell you.
Answers
- \displaystyle \frac{1}{2}. Divide by n^3: \displaystyle \frac{2 + 5/n^3}{4 - 1/n} \to \frac{2}{4}.
- \displaystyle -\frac{4}{3}. Divide by n: \displaystyle \frac{7/n - 4}{3 + 1/n} \to -\frac{4}{3}.
- Diverges to \infty. Divide by n: \displaystyle \frac{n + 1/n}{2 + 3/n}; the top grows without bound and the bottom tends to 2.
- \displaystyle \frac{3}{4}. The conjugate gives \displaystyle \frac{3n}{\sqrt{4n^2 + 3n} + 2n} = \frac{3}{\sqrt{4 + 3/n} + 2} \to \frac{3}{2 + 2}.
- \displaystyle \frac{1}{3}. Divide by 3^n: \displaystyle \frac{1 + 2(1/3)^n}{3 - (1/3)^n} \to \frac{1}{3}, since (1/3)^n \to 0.
- 4. \displaystyle \left|-\frac{2}{3}\right| < 1, so \displaystyle \left(-\frac{2}{3}\right)^n \to 0.
- Diverges. The even terms \displaystyle \frac{n}{n+1} tend to 1 and the odd terms to -1, so far out they lie above \displaystyle \frac{3}{4} and below \displaystyle -\frac{3}{4}. No number is within \displaystyle \frac{1}{2} of both, so taking \displaystyle \varepsilon = \frac{1}{2} defeats every proposed limit.
- \ln 3. The difference is \displaystyle \ln \frac{3n + 1}{n} = \ln\left(3 + \frac{1}{n}\right), and \ln is continuous at 3.
- e^{-6} \approx 0.00248. With \displaystyle t = \frac{1}{x}, \displaystyle 3x \ln\left(1 - \frac{2}{x}\right) = 3 \cdot \frac{\ln(1 - 2t)}{t} \to 3 \cdot (-2), the difference quotient of \ln(1 - 2t) at 0 tending to its derivative -2.
- 1. With \displaystyle t = \frac{1}{n} \to 0^+, \displaystyle n \sin \frac{1}{n} = \frac{\sin t}{t} \to 1 (The squeeze theorem and the limit of sine x over x).
- 1. \displaystyle \ln n^{1/n} = \frac{\ln n}{n}, and \displaystyle \frac{\ln x}{x} \to 0 as x \to \infty (Comparing growth rates at infinity), so n^{1/n} \to e^0 = 1.
- 0. \displaystyle -\frac{1}{\sqrt{n}} \le \frac{\sin(n^2)}{\sqrt{n}} \le \frac{1}{\sqrt{n}}, and both walls tend to 0.
- 4. Divide by n: \displaystyle \frac{4 - \sin(3n)/n}{1 + 2/n}. By the squeeze, \displaystyle \frac{\sin(3n)}{n} \to 0, so the limit is \displaystyle \frac{4}{1}.
- 0. For n \ge 2, \displaystyle \frac{2^n}{n!} = \frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3} \cdots \frac{2}{n} \le 2 \cdot \frac{2}{n}, since the middle factors are at most 1. So \displaystyle 0 < a_n \le \frac{4}{n}, and the squeeze gives 0.
- N = 4997. \displaystyle |a_n - 2| = \left|\frac{2n + 1 - 2n - 6}{n + 3}\right| = \frac{5}{n + 3}, which is less than \varepsilon exactly when \displaystyle n > \frac{5}{\varepsilon} - 3. Given \varepsilon > 0, \displaystyle N = \frac{5}{\varepsilon} - 3 works. For \varepsilon = 0.001, n > 4997; at n = 4997 the distance is exactly \displaystyle \frac{5}{5000} = 0.001, not less.
- 3. \displaystyle a_n = 3 - \frac{12}{n + 4}, and \displaystyle \frac{12}{n + 4} decreases, so a_n increases. Bounds: \displaystyle \frac{3}{5} = a_1 \le a_n < 3. The limit is 3 - 0 = 3.
- 0. \displaystyle \frac{a_{n+1}}{a_n} = \frac{n + 1}{2n} \le 1, with equality only at n = 1, so \displaystyle a_1 = a_2 = \frac{1}{2} and the terms decrease after that; all are positive. By monotone convergence a_n \to L, and \displaystyle a_{n+1} = \frac{n+1}{2n} a_n gives \displaystyle L = \frac{1}{2}L, so L = 0.
- 0. \displaystyle \frac{a_{n+1}}{a_n} = \frac{5}{n + 1}, which is more than 1 for n \le 3, equal to 1 at n = 4, and less than 1 for n \ge 5. The terms rise to \displaystyle a_4 = a_5 = \frac{625}{24} \approx 26.04, then decrease, staying positive. Nonincreasing from n = 4 on and bounded below, the sequence converges; \displaystyle a_{n+1} = \frac{5}{n+1} a_n gives L = 0 \cdot L = 0.
- Limit 400; first above 390 at dose 11. If a_n < 400 then a_{n+1} < 280 + 120 = 400, and a_{n+1} - a_n = 120 - 0.3a_n > 0. The limit solves L = 0.7L + 120. The gap gives a_n = 400 - 280(0.7)^{n-1}, with a_{10} \approx 388.70 and a_{11} \approx 392.09.
- 160. If a_n > 160 then a_{n+1} > 80 + 80 = 160, and a_{n+1} - a_n = 80 - 0.5a_n < 0. Decreasing and bounded below by 160, it converges, and L = 0.5L + 80 gives L = 160. In closed form a_n = 160 + 140(0.5)^{n-1}.
- 3. If 0 < a_n < 3 then \displaystyle 0 < a_{n+1} < \sqrt{9} = 3; and a_{n+1} > a_n exactly when 3a_n > a_n^2, true for 0 < a_n < 3. The limit solves \displaystyle L = \sqrt{3L}, so L = 0 or L = 3, and L \ge a_1 = 1 rules out 0. In closed form a_n = 3^{1 - 2^{1-n}}.
- \displaystyle \sqrt{3}. The terms are \displaystyle 1, 2, \frac{7}{4}, \frac{97}{56}, \ldots, all positive. Since \displaystyle a_{n+1} - \sqrt{3} = \frac{(a_n - \sqrt{3})^2}{2a_n} \ge 0, every term from a_2 on is at least \displaystyle \sqrt{3}. Then \displaystyle a_{n+1} - a_n = \frac{3 - a_n^2}{2a_n} \le 0 for n \ge 2. The limit solves \displaystyle L = \frac{1}{2}\left(L + \frac{3}{L}\right), so L^2 = 3, and \displaystyle L \ge \sqrt{3} gives \displaystyle L = \sqrt{3}. Already \displaystyle \frac{97}{56} \approx 1.73214.
- (-1)^n is bounded and diverges; a_n = n is increasing and diverges. Together the hypotheses give the monotone convergence theorem: a sequence that is both monotone and bounded converges.
- By the reverse triangle inequality, \big||a_n| - |L|\big| \le |a_n - L|, so whatever N works for a_n \to L with a given \varepsilon also works for |a_n| \to |L|. Example: a_n = (-1)^n diverges while |a_n| = 1 converges. When L = 0, |a_n - 0| = \big||a_n| - 0\big|, so the two statements are the same inequality.
- Neither: the terms 2, 4, 16, 256, \ldots grow without bound, so the sequence diverges to infinity. The equation says only that if the sequence converges (and f(x) = x^2 is continuous), its limit is a root. It cannot show convergence, and it does not say which root, if any, is the limit.