Direct comparison test: Worked examples — Maths deep dive

The whole game of the Direct Comparison Test (DCT) is: compare your unknown series, term by term, to a series whose fate you already know. Two benchmarks do almost all the work:

Recall The two benchmark families (from the parent note)

$p$ -series ::: $\sum \frac{1}{n ^{p}}$ converges when $p > 1$ , diverges when $p \leq 1$ . See p-series test. Geometric series ::: $\sum r^{n}$ converges when $∣ r ∣ < 1$ , diverges when $∣ r ∣ \geq 1$ . See Geometric series. Harmonic series ::: $\sum \frac{1}{n}$ — the borderline case ( $p = 1$ ), diverges. See Harmonic series.

The scenario matrix

Every DCT problem falls into one of these cells. We will hit each with at least one worked example.

Cell	Situation	What makes it tricky	Example
A	Denominator slightly bigger than a $p$ -series	direction: trap from above	Ex 1
B	Denominator slightly smaller than harmonic	direction: push from below	Ex 2
C	Bounded wiggling numerator ( $sin$ , $cos$ )	replace numerator by its max/min	Ex 3
D	Geometric-type terms ( $r^{n}$ )	benchmark is geometric, not $p$ -series	Ex 4
E	Extra positive term in denominator makes it smaller	naive direction is the useless one	Ex 5
F	Sign-changing terms	DCT needs $a_{n} \geq 0$ — must repair first	Ex 6
G	Degenerate / limiting input (term $\to$ constant)	test cannot apply — spot it	Ex 7
H	Word problem (real-world total)	translate first, then compare	Ex 8
I	Exam twist: comparison chosen wrong way	steel-man the mistake, then fix	Ex 9

The one rule that decides everything:

The figure above is your compass for the whole page. On the left, a convergent series $\sum b_{n}$ has a finite total; call that finite total $B$ — the green dashed roof. A series trapped underneath it (mint bars) can never sum past $B$ → converges. On the right, a divergent series builds an infinite floor (butter bars) that climbs forever; a series pushed above it (coral bars) is dragged up too → diverges. Every example below is one of these two pictures.

Cell A — denominator slightly bigger than a $p$ -series

Worked example Example 1 —

$n = 1 \sum \infty \frac{1}{n ^{3} + 2 n}$ (Cell A) Here the general term is $a_{n} = \frac{1}{n ^{3} + 2 n}$ . Forecast: the dominant power downstairs is $n^{3}$ , so it should behave like $\sum 1/ n^{3}$ (a $p$ -series with $p = 3 > 1$ ). Guess: converges. Which direction do we need? To prove convergence we need a bigger convergent partner.

Step 1. Note $n^{3} + 2 n > n^{3}$ for all $n \geq 1$ . Why this step? Adding the positive quantity $2 n$ makes the denominator bigger, which makes the whole fraction smaller.

Step 2. Flip the inequality when we take reciprocals of positive numbers: $0 \leq a_{n} = \frac{1}{n ^{3} + 2 n} < \frac{1}{n ^{3}} .$ Why this step? For positive $x < y$ we have $\frac{1}{x} > \frac{1}{y}$ ; reciprocation reverses the order. Now we have our unknown series trapped under a bigger series.

Step 3. Benchmark: $\sum \frac{1}{n ^{3}}$ is a $p$ -series with $p = 3 > 1$ → converges (p-series test). Why this step? We need a partner whose fate we already know for sure; the $p$ -series test hands us that verdict instantly.

Step 4. Conclusion: $0 \leq a_{n} \leq \frac{1}{n ^{3}}$ and the roof converges → our series converges. ✅

Verify: the partial sum $\sum_{n = 1}^{10} \frac{1}{n ^{3} + 2 n} \approx 0.4789$ and it must stay below $\sum_{n = 1}^{10} \frac{1}{n ^{3}} \approx 1.1975$ . It does — the bars sit under the roof, confirming the trap.

Cell B — denominator slightly smaller than harmonic

Worked example Example 2 —

$n = 1 \sum \infty \frac{1}{n + 3}$ (Cell B) Here $a_{n} = \frac{1}{n + 3}$ . Forecast: downstairs grows like $n = n^{1/2}$ , so terms behave like $\sum 1/ n^{1/2}$ , a $p$ -series with $p = \frac{1}{2} \leq 1$ → diverges. To prove divergence we need a smaller divergent partner.

Step 1. We want a smaller fraction than ours, so we enlarge the denominator in a clean, controlled way. For every $n \geq 1$ we have $n \geq 1$ , hence $3 \leq 3 n$ , and therefore $n + 3 \leq n + 3 n = 4 n .$ Why this step? Replacing the awkward $n + 3$ by the larger, tidy $4 n$ makes the fraction smaller yet leaves a recognizable $p$ -series partner. Note the bound holds for all $n \geq 1$ , so no "eventually" caveat is needed.

Step 2. Reciprocate (bigger denominator → smaller fraction): $a_{n} = \frac{1}{n + 3} \geq \frac{1}{4 n} = \frac{1}{4} \cdot \frac{1}{n ^{1/2}} \geq 0.$ Why this step? This points the inequality the right way for divergence — our series sits above the partner.

Step 3. Benchmark: $\sum \frac{1}{4} \cdot \frac{1}{n ^{1/2}} = \frac{1}{4} \sum n^{- 1/2}$ is a constant times a $p$ -series with $p = \frac{1}{2} \leq 1$ → diverges. Why this step? A nonzero constant never changes convergence status, so this floor is genuinely divergent.

Step 4. Our series sits above a divergent floor → diverges. ✅

Verify: at $n = 100$ , $\frac{1}{100 + 3} = \frac{1}{13} \approx 0.0769$ and $\frac{1}{4} \cdot \frac{1}{100} = \frac{1}{40} = 0.025$ . Indeed $0.0769 \geq 0.025$ — floor confirmed.

Cell C — bounded wiggling numerator

Worked example Example 3 —

$n = 1 \sum \infty \frac{3 + cos ( 2 n )}{n ^{2}}$ (Cell C) Here $a_{n} = \frac{3 + cos ( 2 n )}{n ^{2}}$ . Forecast: the numerator wiggles but stays bounded; the $n^{2}$ dominates → should behave like $\sum 1/ n^{2}$ ( $p = 2 > 1$ ) → converges. Prove convergence → need a bigger convergent partner.

Step 1. Bound the wiggle: since $- 1 \leq cos (2 n) \leq 1$ , we have $2 \leq 3 + cos (2 n) \leq 4$ . Why this step? We can't sum $cos$ directly, so we replace the numerator by its maximum $4$ to lose the mess but keep a clean upper bound.

Step 2. Divide by $n^{2} > 0$ (preserves the inequality): $0 \leq a_{n} = \frac{3 + c o s ( 2 n )}{n ^{2}} \leq \frac{4}{n ^{2}} .$ Why this step? The lower bound $2 > 0$ guarantees non-negativity (DCT requires it), and the upper bound gives our roof.

Step 3. Benchmark: $\sum \frac{4}{n ^{2}} = 4 \sum \frac{1}{n ^{2}}$ converges ( $p = 2 > 1$ ). Why this step? The constant $4$ doesn't affect convergence, so this is a certified finite roof.

Step 4. Trapped under a finite roof → converges. ✅

Verify: $\sum_{n = 1}^{20} \frac{3 + c o s ( 2 n )}{n ^{2}} \approx 3.9036$ , and it must sit below $\sum_{n = 1}^{20} \frac{4}{n ^{2}} \approx 6.1854$ . It does.

Cell D — geometric-type terms

Worked example Example 4 —

$n = 1 \sum \infty \frac{2 ^{n}}{3 ^{n} + 1}$ (Cell D) Here $a_{n} = \frac{2 ^{n}}{3 ^{n} + 1}$ . Forecast: this is not a $p$ -series — the variable is in the exponent. The ratio $2^{n} / 3^{n} = (2/3)^{n}$ with $∣ r ∣ = 2/3 < 1$ suggests a geometric benchmark → converges. Prove convergence → bigger convergent partner.

Step 1. Drop the $+ 1$ to shrink the denominator, which enlarges the fraction: $a_{n} = \frac{2 ^{n}}{3 ^{n} + 1} < \frac{2 ^{n}}{3 ^{n}} = (\frac{2}{3})^{n} .$ Why this step? A smaller denominator gives a bigger fraction — exactly the direction we need for an upper trap. And $3^{n}$ alone is a clean geometric term.

Step 2. Confirm non-negativity: $2^{n} > 0$ and $3^{n} + 1 > 0$ , so every term is $\geq 0$ . Good.

Step 3. Benchmark: $\sum (2/3)^{n}$ is geometric with $r = \frac{2}{3}$ , $∣ r ∣ < 1$ → converges. Why this step? The geometric-series test gives an immediate, certain verdict for terms with the index in the exponent — the right benchmark family here, not the $p$ -series family.

Step 4. Trapped under a convergent geometric roof → converges. ✅

Verify: the tail decays like $(2/3)^{n}$ . Numerically $\sum_{n = 1}^{\infty} (2/3)^{n} = \frac{2/3}{1 - 2/3} = 2$ , and $\sum_{n = 1}^{15} \frac{2 ^{n}}{3 ^{n} + 1} \approx 1.2494 < 2$ — under the roof.

Cell E — extra positive term makes it smaller (trap for the arrow)

Worked example Example 5 —

$n = 1 \sum \infty \frac{n}{n ^{2} + 1}$ (Cell E) Here $a_{n} = \frac{n}{n ^{2} + 1}$ . Forecast: downstairs $n^{2}$ , upstairs $n$ , so terms behave like $\frac{n}{n ^{2}} = \frac{1}{n}$ → harmonic-like → diverges. Prove divergence → smaller divergent partner. Beware: the " $+ 1$ " makes the fraction smaller than $\frac{1}{n}$ , so the obvious comparison points the useless way.

Step 1. First observe the trap. $n^{2} + 1 > n^{2} \Rightarrow \frac{n}{n ^{2} + 1} < \frac{n}{n ^{2}} = \frac{1}{n}$ . That is $a_{n} < \frac{1}{n}$ — a smaller divergent partner tells us nothing. Why this step? Recognizing the useless direction first saves you from a wrong conclusion (see the mistake in the parent note).

Step 2. Repair: bound the denominator by something larger than $n^{2} + 1$ but still $\sim n^{2}$ . For $n \geq 1$ , $n^{2} + 1 \leq n^{2} + n^{2} = 2 n^{2}$ . Why this step? Enlarging the denominator to $2 n^{2}$ makes the fraction smaller in a controlled way that still leaves a divergent partner.

Step 3. Then $a_{n} = \frac{n}{n ^{2} + 1} \geq \frac{n}{2 n ^{2}} = \frac{1}{2} \cdot \frac{1}{n} \geq 0.$ Why this step? Now our series sits above $\frac{1}{2} \sum \frac{1}{n}$ , a divergent floor.

Step 4. Benchmark: $\frac{1}{2} \sum \frac{1}{n}$ is half the harmonic series → diverges. Why this step? Halving every term of a divergent series keeps it divergent, so the floor is genuinely infinite.

Step 5. Pushed above an infinite floor → diverges. ✅

Verify: at $n = 50$ , $\frac{50}{2501} \approx 0.01999$ and $\frac{1}{2} \cdot \frac{1}{50} = 0.01$ . Indeed $0.01999 \geq 0.01$ — floor holds.

Cell F — sign-changing terms (DCT can't run raw)

Worked example Example 6 —

$n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n ^{2}}$ (Cell F) Here $a_{n} = \frac{( - 1 ) ^{n}}{n ^{2}}$ . Forecast: terms alternate in sign, so DCT — which demands $a_{n} \geq 0$ — cannot be applied directly. But we can compare absolute values.

Step 1. Recognize the obstacle: $\frac{( - 1 ) ^{n}}{n ^{2}}$ is sometimes negative, so partial sums are not monotone increasing and Monotone Convergence fails. Raw DCT is illegal. Why this step? Naming why the test breaks is half the skill — this is the exact failure the parent note warns about.

Step 2. Compare the absolute value instead: $∣ a_{n} ∣ = \frac{( - 1 ) ^{n}}{n ^{2}} = \frac{1}{n ^{2}} \leq \frac{1}{n ^{2}} .$ Why this step? $∣ a_{n} ∣ \geq 0$ , so DCT is legal on absolute values. Here $∣ a_{n} ∣ = \frac{1}{n ^{2}}$ is a convergent $p$ -series ( $p = 2 > 1$ ).

Step 3. Since $\sum ∣ a_{n} ∣$ converges, the original series converges absolutely → hence converges (Absolute convergence).

Step 4. Conclusion: converges (absolutely). ✅

Verify: $\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n}}{n ^{2}} = - \frac{π ^{2}}{12} \approx - 0.8225$ , a finite value — convergence confirmed. And $\sum ∣ a_{n} ∣ = \frac{π ^{2}}{6} \approx 1.6449$ is finite, confirming absolute convergence.

Cell G — degenerate / limiting input (test cannot apply)

Worked example Example 7 —

$n = 1 \sum \infty \frac{n}{n + 1}$ (Cell G) Here $a_{n} = \frac{n}{n + 1}$ . Forecast: watch the terms as $n \to \infty$ : $\frac{n}{n + 1} \to 1$ , not $0$ . A series whose terms don't shrink to $0$ cannot converge — no comparison needed.

Step 1. Compute the limiting value: $n \to \infty lim \frac{n}{n + 1} = n \to \infty lim \frac{1}{1 + 1/ n} = 1$ . Why this step? The $n$ -th term test (a prerequisite of every comparison): if $a_{n} \neq \to 0$ , the series diverges instantly. DCT is unnecessary and would be wasteful here.

Step 2. Since terms approach $1 \neq = 0$ , the running total grows without bound. Why this step? You are adding numbers each near $1$ forever — the sum climbs like $\approx n$ .

Step 3. Conclusion: diverges by the term test. ✅ (DCT is the wrong tool — spotting that is the lesson.)

Verify: $\sum_{n = 1}^{100} \frac{n}{n + 1} \approx 94.808$ , already near $100$ and climbing linearly — clearly unbounded.

Cell H — a real-world word problem

Worked example Example 8 — Ink on an infinite ruler

(Cell H) A printer marks a ruler: at position $n$ (cm) it prints a line whose ink area is $\frac{1}{n ^{2} + n}$ mm². Does the total ink used over infinitely many marks stay finite?

Here $a_{n} = \frac{1}{n ^{2} + n}$ . Forecast: areas shrink like $1/ n^{2}$ → the total should be finite (converges). Prove convergence → bigger convergent partner.

Step 1. Translate: total ink $= n = 1 \sum \infty \frac{1}{n ^{2} + n}$ , all terms $> 0$ (areas). Good — DCT legal. Why this step? Physical quantities (area, mm²) are non-negative, so the sign condition is automatic.

Step 2. Bound below the roof: $n^{2} + n > n^{2} \Rightarrow \frac{1}{n ^{2} + n} < \frac{1}{n ^{2}}$ . Why this step? Bigger denominator → smaller fraction → we sit under $\frac{1}{n ^{2}}$ .

Step 3. Benchmark $\sum \frac{1}{n ^{2}}$ converges ( $p = 2 > 1$ ) → total ink is finite. Why this step? A finite roof means the ink jar never overflows.

Step 4. Conclusion: total ink is finite (converges). ✅ Units: each term is mm², a sum of areas → total in mm².

Verify: in fact $\frac{1}{n ^{2} + n} = \frac{1}{n} - \frac{1}{n + 1}$ telescopes to exactly $\sum_{n = 1}^{\infty} = 1$ mm². It converges, and $1 < \frac{π ^{2}}{6} \approx 1.645$ — under the $p$ -series roof, as promised.

Cell I — the exam twist (wrong-arrow steel-man)

Worked example Example 9 — Spot & fix the flawed argument:

$n = 1 \sum \infty \frac{1}{2 n}$ (Cell I) Here $a_{n} = \frac{1}{2 n}$ . A student writes: "Since $\frac{1}{2 n} \leq \frac{1}{n}$ and $\sum \frac{1}{n}$ ... converges? — so $\sum \frac{1}{2 n}$ converges." Find the two errors.

Forecast: $\frac{1}{2 n} = \frac{1}{2} \cdot \frac{1}{n}$ is half the harmonic series → diverges. The student's argument is broken twice.

Step 1. Error 1 (false benchmark): $\sum \frac{1}{n}$ is the harmonic series and it diverges, not converges. Why this step? Comparing to a series whose behaviour you've mis-remembered poisons everything downstream (Harmonic series).

Step 2. Error 2 (wrong arrow): even if we knew $\sum \frac{1}{n}$ diverges, showing $\frac{1}{2 n} \leq \frac{1}{n}$ puts our series below a divergent one — the useless direction. Being smaller than a divergent series proves nothing. Why this step? This is the classic "arrow mismatch": divergence must be shown from below (smaller divergent partner).

Step 3. Correct comparison — build the RIGHT inequality. We need a smaller divergent partner sitting below our series. Take $b_{n} = \frac{1}{2 n + 2}$ . Since $2 n + 2 > 2 n$ we have $a_{n} = \frac{1}{2 n} > \frac{1}{2 n + 2} = \frac{1}{2} \cdot \frac{1}{n + 1} \geq 0.$ Why this step? Now our series is genuinely above a partner, and that partner is $\frac{1}{2} \sum \frac{1}{n + 1}$ — a shifted half-harmonic series.

Step 4. Benchmark the partner. $\frac{1}{2} \sum_{n = 1}^{\infty} \frac{1}{n + 1} = \frac{1}{2} \sum_{m = 2}^{\infty} \frac{1}{m}$ is the harmonic series (missing only its first term) times $\frac{1}{2}$ → diverges. Why this step? Dropping one term and halving the rest cannot rescue a divergent series, so the floor is truly infinite.

Step 5. Conclusion. $\frac{1}{2 n} \geq \frac{1}{2 n + 2} \geq 0$ and the smaller partner diverges → $\sum \frac{1}{2 n}$ diverges. ✅ (Equivalently, $\sum \frac{1}{2 n} = \frac{1}{2} \sum \frac{1}{n}$ is a nonzero constant times the harmonic series, which diverges.)

Verify: $\sum_{n = 1}^{N} \frac{1}{2 n} = \frac{1}{2} H_{N}$ where $H_{N} \to \infty$ ; e.g. $\frac{1}{2} H_{1000} \approx 3.742$ and growing like $\frac{1}{2} ln N$ — unbounded.

Recall checkpoint

Recall Quick self-test

On this page, what does the symbol $a_{n}$ always stand for? ::: The general term of the series being tested — the exact expression inside the summation sign. To trap $\sum \frac{1}{n ^{3} + 2 n}$ we compared to which convergent series, and via what inequality? ::: $\frac{1}{n ^{3} + 2 n} < \frac{1}{n ^{3}}$ (bigger denominator → smaller fraction); $\sum \frac{1}{n ^{3}}$ converges. Why can't we run DCT directly on $\sum \frac{( - 1 ) ^{n}}{n ^{2}}$ ? ::: Terms change sign, so partial sums aren't monotone; compare absolute values instead (absolute convergence). $\sum \frac{n}{n + 1}$ — what one-line test kills it? ::: Terms $\to 1 \neq = 0$ , so the $n$ -th term test forces divergence; no comparison needed. In Example 5, why is $\frac{n}{n ^{2} + 1} < \frac{1}{n}$ the useless direction? ::: It shows our series is smaller than a divergent series, which proves nothing; we repaired it with $\leq 2 n^{2}$ to get a smaller divergent floor $\frac{1}{2 n}$ . For a bounded-numerator series $\frac{3 + c o s 2 n}{n ^{2}}$ , what value replaces the numerator to get a convergent roof? ::: Its maximum, $4$ , giving $\leq \frac{4}{n ^{2}}$ .

Connections

Direct comparison test — the parent rule these examples exercise.
p-series test — benchmark for Cells A, B, C, E, H.
Geometric series — benchmark for Cell D.
Harmonic series — benchmark for Cells B, E, I.
Absolute convergence — the repair for sign-changing Cell F.
Limit comparison test — smoother alternative when the algebra in Cells B/E gets ugly.
Integral test — another way to classify the $p$ -series benchmarks.
Monotone convergence theorem — the engine, and the reason Cell F fails.