Sequences — convergence, divergence, boundedness, monotonicity: Worked examples

The scenario matrix

Every sequence you meet falls into one of these case classes. Each row is a distinct trap or technique; the worked examples below are tagged with the cell they cover so you can see the whole space is filled.

#	Case class	What makes it tricky	Weapon of choice	Example
A	Rational in $n$ (polynomial over polynomial)	which power dominates?	divide by top power of $n$	Ex 1
B	Difference of two things that both $\to \infty$	$\infty - \infty$ is undefined	rationalise / rewrite	Ex 2
C	Trapped oscillation ( $sin$ , $cos$ in numerator)	numerator never settles	Squeeze Theorem	Ex 3
D	Pure oscillation (no shrinking)	no limit at all	$ε$ – $N$ contradiction	Ex 4
E	Factorial / exponential race	which grows faster?	ratio $a_{n + 1} / a_{n}$	Ex 5
F	Recursive (defined by itself)	limit hidden in a fixed point	MCT + solve $L = f (L)$	Ex 6
G	The number $e$ shape $(1 + x / n)^{n}$	looks like $1^{\infty}$	known $e$ -limit + logs	Ex 7
H	Zero / degenerate & sign cases	$n = 0$ ?, negative terms, $r = 1$ ?	check every value of a parameter	Ex 8
I	Real-world word problem	translate words → sequence	model then take limit	Ex 9
J	Exam twist (piecewise / hidden divergence)	a "nice" formula that still diverges	split into subsequences	Ex 10

We now walk every cell.

Case A — rational in $n$

Case B — difference of two infinities ( $\infty - \infty$ )

Case C — oscillation trapped by a shrinking envelope

Worked example

(Cell C) Find $n \to \infty lim \frac{sin n}{n}$ . Forecast: $sin n$ wobbles forever between $- 1$ and $1$ . Dividing by a growing $n$ — does the wobble survive?

Bound the wobble. For every $n$ , $- 1 \leq sin n \leq 1$ , so dividing by the positive number $n$ : $- \frac{1}{n} \leq \frac{s i n n}{n} \leq \frac{1}{n} .$ Why this step? We cannot compute $sin n$ term by term, but we CAN cage it between two sequences we understand. Look at the figure: the sequence lives at isolated dots (there is no term "between" $n = 3$ and $n = 4$ ), each magenta dot sitting inside the violet–orange funnel $\pm \frac{1}{n}$ .
Squeeze. Both $- \frac{1}{n} \to 0$ and $\frac{1}{n} \to 0$ . By the Squeeze Theorem, $\frac{s i n n}{n} \to 0$ . Why this step? If both walls of the funnel close on $0$ , whatever is trapped inside has nowhere to go but $0$ . See Squeeze Theorem.

Verify: $n = 50$ : $\frac{s i n 50}{50} = \frac{- 0.2624}{50} = - 0.00525$ , well within $\pm \frac{1}{50} = \pm 0.02$ . ✓

Case D — pure oscillation, no shrinking

Worked example

(Cell D) Show $a_{n} = cos (nπ)$ diverges. Forecast: $cos (nπ)$ equals... what for $n = 1, 2, 3, 4$ ? Work it out, then guess.

Simplify. $cos (0) = 1, cos (π) = - 1, cos (2 π) = 1, \dots$ so $a_{n} = (- 1)^{n}$ : the sequence $- 1, 1, - 1, 1, \dots$ Why this step? Recognising the disguise ( $cos (nπ) = (- 1)^{n}$ ) reduces a "trig mystery" to the classic oscillator.
$ε$ – $N$ contradiction. Suppose it converged to some $L$ . Take the tolerance $ε = 1$ . Convergence demands all terms past some $N$ sit inside the band $(L - 1, L + 1)$ , which has width $2$ . But consecutive terms are $1$ and $- 1$ — a distance of exactly $2$ apart. Two points distance $2$ apart cannot both lie strictly inside a band of width $2$ . Contradiction. Why this step? Convergence forces the tail to cluster; a fixed gap of $2$ forbids clustering. No $L$ can work, so it diverges.

Verify: The set of values is exactly ${- 1, 1}$ — two accumulation points, so no single limit. ✓

Case E — factorial vs power race

Worked example

(Cell E) Find $n \to \infty lim \frac{n !}{n ^{n}}$ . Forecast: $n!$ is a giant. But $n^{n}$ is $n$ copies of $n$ multiplied, while $n! = 1 \cdot 2 \dots n$ uses many small factors. Who wins?

Look at ONE typical term. $\frac{n !}{n ^{n}} = \frac{1 \cdot 2 \cdot 3 \dots n}{n \cdot n \cdot n \dots n} = \frac{1}{n} \cdot \frac{2}{n} \dots \frac{n}{n} \leq \frac{1}{n} \cdot 1 \cdot 1 \dots 1 = \frac{1}{n} .$ Why this step? Every factor $\frac{k}{n} \leq 1$ , and the very first factor $\frac{1}{n}$ already drags the whole product down. So $0 < \frac{n !}{n ^{n}} \leq \frac{1}{n}$ .
Squeeze. $0 \to 0$ and $\frac{1}{n} \to 0$ , so the sequence $\to 0$ . Why this step? Same funnel logic as Cell C — caged between $0$ and $\frac{1}{n}$ .
Cross-check with the term-ratio shrink factor. Look at $\frac{a _{n + 1}}{a _{n}} = (\frac{n}{n + 1})^{n} \to e^{- 1} < 1$ . (This is NOT the series ratio test — we are only asking how each term compares to the previous one.) When $\frac{a _{n + 1}}{a _{n}}$ eventually stays below a fixed number $< 1$ , the terms shrink at least geometrically, so they must tend to $0$ . See The number e.

Verify: $n = 10$ : $\frac{10 !}{1 0 ^{10}} = \frac{3 628 800}{1 0 ^{10}} = 0.00036288$ , and $\frac{1}{10} = 0.1$ — safely trapped and tiny. ✓

Case F — recursive sequence

Worked example

(Cell F) Let $a_{1} = 2, a_{n + 1} = \frac{1}{2} (a_{n} + \frac{2}{a _{n}})$ . Find $lim a_{n}$ . Forecast: This is Newton's iteration in disguise. Compute $a_{2}$ by hand and guess where it heads.

Bounded below by $2$ . By AM–GM, $\frac{1}{2} (a_{n} + \frac{2}{a _{n}}) \geq a_{n} \cdot \frac{2}{a _{n}} = 2$ , so $a_{n + 1} \geq 2$ for all $n \geq 1$ . Why this step? MCT needs a floor. AM–GM ( $\frac{1}{2} (x + y) \geq x y$ ) supplies it because the product $a_{n} \cdot \frac{2}{a _{n}} = 2$ is constant.
Decreasing. $a_{n} - a_{n + 1} = a_{n} - \frac{1}{2} (a_{n} + \frac{2}{a _{n}}) = \frac{a _{n}^{2} - 2}{2 a _{n}} \geq 0$ since $a_{n} \geq 2 \Rightarrow a_{n}^{2} \geq 2$ . Why this step? MCT needs monotonicity; the sign of the difference gives it.
Apply MCT and solve the fixed point. Decreasing + bounded below ⟹ converges to some $L$ . Both $a_{n}$ and $a_{n + 1}$ tend to the same $L$ , so $L = \frac{1}{2} (L + \frac{2}{L}) \Rightarrow 2 L^{2} = L^{2} + 2 \Rightarrow L^{2} = 2 \Rightarrow L = 2$ (positive branch, since terms are $> 0$ ). Why this step? The limit must satisfy the same relation the sequence obeys — that is the fixed-point equation. Rests on Completeness of the Real Numbers (which guarantees the $in f$ exists).

Verify: $a_{2} = \frac{1}{2} (2 + 1) = 1.5$ , $a_{3} = \frac{1}{2} (1.5 + \frac{2}{1.5}) = 1.41 \overline{6}$ , $a_{4} \approx 1.414216$ . Indeed $\to 2 \approx 1.414214$ . ✓

Case G — the $e$ shape $(1 + x / n)^{n}$

Worked example

(Cell G) Find $n \to \infty lim (1 + \frac{3}{n})^{n}$ . Forecast: The base $1 + \frac{3}{n} \to 1$ but the exponent $n \to \infty$ . That is the indeterminate $1^{\infty}$ — could be $1$ , could be $\infty$ . Guess.

Recall the defining $e$ -limit. From The number e, $(1 + \frac{x}{n})^{n} \to e^{x}$ . Why this step? The whole difficulty is $1^{\infty}$ ; the $e$ -limit is the only tool that resolves it cleanly, because $e$ is defined as exactly this kind of limit.
Match $x = 3$ . Directly, $(1 + \frac{3}{n})^{n} \to e^{3}$ . Why this step? Our expression is the template with $x = 3$ — no algebra needed once you recognise the shape.
(Optional log proof, done rigorously) Take logs: $ln a_{n} = n ln (1 + \frac{3}{n})$ . Write $t = \frac{3}{n}$ , so $t \to 0$ as $n \to \infty$ , and $ln a_{n} = 3 \cdot \frac{l n ( 1 + t )}{t}$ . The key fact is the standard limit $t \to 0 lim \frac{ln ( 1 + t )}{t} = 1$ (it is the derivative of $ln$ at $1$ , since $\frac{d}{d x} ln x_{x = 1} = 1$ — see Limits of functions). Therefore $ln a_{n} \to 3 \cdot 1 = 3$ , and since $exp$ is continuous, $a_{n} = e^{l n a_{n}} \to e^{3}$ . Why this step? Taking logs turns the scary exponent into a product $3 \cdot \frac{l n ( 1 + t )}{t}$ ; we then lean on ONE rigorously-established limit rather than an unjustified series truncation.

Verify: $n = 1000$ : $(1 + \frac{3}{1000})^{1000} \approx 20.06$ , and $e^{3} \approx 20.09$ . ✓ Closing in from below (convergence to $e^{x}$ is slow, so a small gap at $n = 1000$ is expected).

Case H — zero / degenerate & sign cases

Worked example

(Cell H) For the geometric sequence $a_{n} = r^{n}$ , classify convergence for every real value of $r$ . Forecast: You know $r = \frac{1}{2}$ shrinks to $0$ and $r = 2$ blows up. But what about $r = - 1$ ? $r = 1$ ? $r = 0$ ? $r = - 2$ ? Enumerate before reading.

Work through every region of the number line (the figure colours each region):

$∣ r ∣ < 1$ (e.g. $r = \frac{1}{2}$ or $r = - \frac{1}{2}$ ): $∣ r^{n} ∣ = ∣ r ∣^{n} \to 0$ , so $a_{n} \to 0$ . The negative case still $\to 0$ because the magnitude shrinks even though signs alternate. Why: a magnitude below $1$ , multiplied by itself repeatedly, vanishes.
$r = 1$ : every term is $1$ , so $a_{n} \to 1$ (converges — the degenerate constant case). Why this matters: $r = 1$ is the knife-edge; the formula $\frac{a}{1 - r}$ for sums would divide by zero here, so it's a special value you must call out.
$r = 0$ : $0^{n} = 0$ for all $n \geq 1$ , so $a_{n} \to 0$ (the zero input — trivially convergent).
$r = - 1$ : $(- 1)^{n}$ oscillates $- 1, 1, - 1, \dots$ — diverges (Cell D).
$r > 1$ (e.g. $r = 2$ ): $r^{n} \to + \infty$ — diverges to infinity.
$r < - 1$ (e.g. $r = - 2$ ): magnitude $\to \infty$ AND sign alternates, giving $- 2, 4, - 8, 16, \dots$ — diverges (unbounded oscillation).

Summary: $r^{n}$ converges exactly when $- 1 < r \leq 1$ (limit $0$ if $∣ r ∣ < 1$ , limit $1$ if $r = 1$ ); diverges otherwise.

Verify: $(0.5)^{20} = 0.00000095$ (→0), $(- 0.5)^{20} = + 0.00000095$ (→0), $1^{20} = 1$ , $(- 1)^{20} = 1$ but $(- 1)^{21} = - 1$ (no limit), $2^{20} = 1048576$ (→∞). ✓

Case I — real-world word problem

Worked example

(Cell I) A drug leaves the body such that each hour $30%$ of the amount present is cleared, and a fresh $10$ mg dose is added. If $a_{n}$ is the mg present just after the $n$ -th dose (with $a_{1} = 10$ ), does the drug level stabilise, and at what value? Forecast: Add-then-decay repeatedly — does it climb forever, or settle at a steady "therapeutic" level? Guess a ceiling.

Model it. After a dose, $70%$ survives to the next hour, then $10$ mg is added: $a_{n + 1} = 0.7 a_{n} + 10$ . Why this step? " $30%$ cleared" means $70% = 0.7$ remains; the $+ 10$ is the new dose. Words → recurrence.
Guess the steady state (fixed point). If it stabilises at $L$ , then $L = 0.7 L + 10 \Rightarrow 0.3 L = 10 \Rightarrow L = \frac{10}{0.3} = 33. \overline{3}$ mg. Why this step? At equilibrium the level stops changing, so $a_{n + 1} = a_{n} = L$ — the same fixed-point trick as Cell F.
Confirm convergence. Subtract the fixed point: $a_{n + 1} - L = 0.7 (a_{n} - L)$ , so the gap shrinks by factor $0.7$ each step: $a_{n} - L = (0.7)^{n - 1} (a_{1} - L) \to 0$ since $∣0.7∣ < 1$ (Cell H). Hence $a_{n} \to L$ . Why this step? The distance to the ceiling is itself a geometric sequence with ratio $0.7$ — and we proved those converge to $0$ .

Verify: $a_{1} = 10, a_{2} = 0.7 (10) + 10 = 17, a_{3} = 0.7 (17) + 10 = 21.9, a_{4} = 25.33$ , climbing toward $33. \overline{3}$ mg. Units: mg throughout — consistent. ✓

Case J — exam twist (hidden divergence)

Worked example

(Cell J) Does $a_{n} = \frac{( - 1 ) ^{n} n}{n + 1}$ converge? Forecast: $\frac{n}{n + 1} \to 1$ , which is nice and settled. Tempting to say the whole thing converges. Look harder before you commit.

Strip the magnitude. $∣ a_{n} ∣ = \frac{n}{n + 1} = \frac{1}{1 + \frac{1}{n}} \to 1$ , so the size of the terms settles near $1$ — they do not shrink. Why this step? Before hunting for a limit, check whether the terms even get small. Here they hover near size $1$ , which is a red flag: a term-magnitude that stays near $1$ can only converge if the sign also settles.
Split into subsequences by parity. Look at even and odd indices separately, because the factor $(- 1)^{n}$ is $+ 1$ on evens and $- 1$ on odds: $even n : a_{n} = + \frac{n}{n + 1} \to + 1, odd n : a_{n} = - \frac{n}{n + 1} \to - 1.$ Why this step? A convergent sequence forces every subsequence to share its one limit. If we exhibit two subsequences with different limits, no single $L$ can exist.
Seal it with an $ε$ – $N$ argument. Suppose $a_{n} \to L$ . Take $ε = \frac{1}{2}$ . Past some $N$ all terms lie in $(L - \frac{1}{2}, L + \frac{1}{2})$ , a band of width $1$ . But for large $n$ an even term is near $+ 1$ and the next (odd) term is near $- 1$ — a distance close to $2$ apart, which cannot fit in a width- $1$ band. Contradiction. Why this step? This is the same clustering argument as Cell D, now dressed in a rational disguise — it converts "two subsequence limits" into a hard contradiction.
Conclude: the two limit points $+ 1$ and $- 1$ mean $a_{n}$ diverges by oscillation.

Verify: $a_{100} = + \frac{100}{101} = 0.9901$ , $a_{101} = - \frac{101}{102} = - 0.9902$ — consecutive terms nearly $2$ apart, forever, so no single limit. ✓

The matrix, revisited

Recall Which weapon for which cell? (test yourself)

Rational $\frac{p ( n )}{q ( n )}$ ::: divide by highest power of $n$ ; compare degrees. Difference $n + 1 - n$ (an $\infty - \infty$ ) ::: multiply by the conjugate to kill the roots. $\frac{s i n n}{n}$ or trapped oscillation ::: Squeeze between $\pm \frac{1}{n}$ . Pure oscillation $(- 1)^{n}$ , $cos (nπ)$ ::: diverges — $ε$ – $N$ contradiction (fixed gap). Factorial vs power $\frac{n !}{n ^{n}}$ ::: bound one term, then squeeze; or term-ratio shrink factor. Recursive $a_{n + 1} = f (a_{n})$ ::: prove monotone + bounded (MCT), then solve $L = f (L)$ . $(1 + \frac{x}{n})^{n}$ shape ::: it is $e^{x}$ by the defining limit. Geometric $r^{n}$ ::: converges iff $- 1 < r \leq 1$ ; check $r = \pm 1, 0$ separately. Word problem "decay + dose" ::: model as $a_{n + 1} = r a_{n} + c$ , fixed point $L = \frac{c}{1 - r}$ . Rational $\times (- 1)^{n}$ ::: split into even/odd subsequences; different limits ⟹ diverges.

See also the Bolzano–Weierstrass Theorem (every bounded sequence has a convergent subsequence — the escape hatch when monotonicity fails), Cauchy Sequences (convergence without naming the limit), Limits of functions (the continuous cousin of these ideas), and Series — convergence tests (where these term-limits become the first thing you check).