Limit comparison test: Worked examples — Maths deep dive

This deep dive is the training ground for the Limit comparison test. The parent note told you the rule and proved it. Here we hunt down every kind of series LCT can be thrown at, so that when the exam hands you a fraction you have never seen, you already know which cell of the table it falls into.

Before we start: recall the one number the whole test lives on, $L = lim_{n \to \infty} \frac{a _{n}}{b _{n}},$ where $a_{n}$ is your series' term and $b_{n}$ is the skeleton you build by keeping the strongest power top and bottom. Read $L$ like a verdict:

$0 < L < \infty$ — same fate, and $\sum b_{n}$ 's fate decides both.
$L = 0$ — $a_{n}$ is smaller; only useful if $\sum b_{n}$ converges.
$L = \infty$ — $a_{n}$ is bigger; only useful if $\sum b_{n}$ diverges.

The scenario matrix

Every LCT problem you will ever meet lives in one of these cells. The last column names the example on this page that nails it.

Cell	What makes it different	Comparison $b_{n}$	Verdict style	Example
A polynomial / polynomial	leading powers, clean $L$	p-Series	main $0 < L < \infty$	Ex 1
B root in denominator	fractional dominant power	$p$ -series with $p = \frac{1}{2}$	main, diverges	Ex 2
C exponential dominates	$2^{n}$ swamps polynomial	Geometric Series	main, converges	Ex 3
D slow log factor	$ln n$ vs power of $n$	$p$ -series, $L = 0$	edge $L = 0$	Ex 4
E wrong-way direct comparison	$L = \infty$ still decides	Harmonic Series	edge $L = \infty$	Ex 5
F degenerate / trap	$a_{n} \neq \to 0$ , or bad $b_{n}$	—	test misused	Ex 6
G real-world word problem	model a total from a rate	$p$ -series	main	Ex 7
H exam twist	two unknown parameters	$p$ -series in $p$	boundary tuning	Ex 8

Cell A — polynomial over polynomial

Worked example Example 1 (Cell A) —

$n = 1 \sum \infty \frac{5 n ^{3} - 2 n + 7}{2 n ^{5} + n ^{2} + 1}$ Forecast: cover the weak terms with your thumb. Top looks like $5 n^{3}$ , bottom like $2 n^{5}$ . Ratio $\sim \frac{5 n ^{3}}{2 n ^{5}} = \frac{5}{2 n ^{2}}$ — that smells like $1/ n^{2}$ , a convergent $p$ -series. Guess: converges.

Pick $b_{n}$ . Keep only the strongest power top and bottom: $b_{n} = \frac{n ^{3}}{n ^{5}} = \frac{1}{n ^{2}} .$ Why this step? For huge $n$ the $- 2 n + 7$ and $+ n^{2} + 1$ are dust next to the leading powers; the ratio of those powers is what governs decay.
Compute $L$ . $L = lim_{n \to \infty} \frac{( 5 n ^{3} - 2 n + 7 ) / ( 2 n ^{5} + n ^{2} + 1 )}{1/ n ^{2}} = lim_{n \to \infty} \frac{n ^{2} ( 5 n ^{3} - 2 n + 7 )}{2 n ^{5} + n ^{2} + 1} .$ Divide top and bottom by $n^{5}$ : $L = lim_{n \to \infty} \frac{5 - 2/ n ^{2} + 7/ n ^{5}}{2 + 1/ n ^{3} + 1/ n ^{5}} = \frac{5}{2} .$ Why this step? Dividing by the highest power $n^{5}$ sends every "weak" term to $0$ , leaving only the leading coefficients.
Conclude. $0 < L = \frac{5}{2} < \infty$ , and $\sum 1/ n^{2}$ converges ( $p = 2 > 1$ ). The series converges. ✔

Verify: $L$ is finite and positive, so $\sum a_{n}$ and $\sum b_{n}$ share a fate; $b_{n}$ 's fate is convergence. Sanity: partial sums of $\sum \frac{5 n ^{3} - 2 n + 7}{2 n ^{5} + n ^{2} + 1}$ settle near a fixed number, consistent with convergence.

Cell B — a root hides a fractional power

Worked example Example 2 (Cell B) —

$n = 1 \sum \infty \frac{n + 1}{n + 4}$ Forecast: top $\sim n = n^{1/2}$ , bottom $\sim n$ . Ratio $\sim n^{1/2} / n = 1/ n^{1/2}$ — a $p$ -series with $p = \frac{1}{2} \leq 1$ . Guess: diverges.

Pick $b_{n} = \frac{1}{n ^{1/2}}$ . Why? The dominant power on top is $n^{1/2}$ , on bottom $n^{1}$ ; their ratio is $n^{1/2 - 1} = n^{- 1/2}$ .
Compute $L$ . $L = lim_{n \to \infty} \frac{( n + 1 ) / ( n + 4 )}{1/ n} = lim_{n \to \infty} \frac{n ( n + 1 )}{n + 4} = lim_{n \to \infty} \frac{n + n}{n + 4} .$ Divide top and bottom by $n$ : $L = lim_{n \to \infty} \frac{1 + 1/ n}{1 + 4/ n} = 1.$ Why this step? Multiplying by $n$ turns the awkward root into a clean polynomial ratio.
Conclude. $0 < L = 1 < \infty$ , and $\sum 1/ n^{1/2}$ diverges ( $p = \frac{1}{2} \leq 1$ ). The series diverges. ✔

Verify: $L = 1$ means the terms behave identically to $1/ n$ for large $n$ ; a Direct Comparison Test proof of divergence would be painful here (the inequalities point the wrong way), which is exactly why LCT is the right tool.

Cell C — exponential beats everything (geometric skeleton)

Worked example Example 3 (Cell C) —

$n = 1 \sum \infty \frac{3 ^{n} + n ^{2}}{4 ^{n} - n}$ Forecast: exponentials crush polynomials. Top $\sim 3^{n}$ , bottom $\sim 4^{n}$ . Ratio $\sim (3/4)^{n}$ — a Geometric Series with $∣ r ∣ = 3/4 < 1$ . Guess: converges.

Look at the figure: the blue $3^{n}$ curve and yellow $4^{n}$ curve both rocket up, but the red ratio $3^{n} / 4^{n}$ collapses to zero geometrically — that collapse is what convergence looks like.

Pick $b_{n} = (\frac{3}{4})^{n} = \frac{3 ^{n}}{4 ^{n}}$ . Why? Keep the dominant exponential top and bottom; $n^{2}$ and $- n$ are negligible beside $3^{n}$ and $4^{n}$ (any exponential with base $> 1$ eventually dwarfs any polynomial).
Compute $L$ . $L = lim_{n \to \infty} \frac{( 3 ^{n} + n ^{2} ) / ( 4 ^{n} - n )}{3 ^{n} / 4 ^{n}} = lim_{n \to \infty} \frac{4 ^{n} ( 3 ^{n} + n ^{2} )}{3 ^{n} ( 4 ^{n} - n )} .$ Split into pieces by dividing top and bottom by $3^{n} 4^{n}$ : $L = lim_{n \to \infty} \frac{1 + n ^{2} / 3 ^{n}}{1 - n / 4 ^{n}} = \frac{1 + 0}{1 - 0} = 1.$ Why this step? $n^{2} / 3^{n} \to 0$ and $n / 4^{n} \to 0$ because exponential growth outpaces polynomial growth — this is the fact the figure shows.
Conclude. $0 < L = 1 < \infty$ , and $\sum (3/4)^{n}$ converges (geometric, $∣ r ∣ < 1$ ). The series converges. ✔

Verify: $\sum_{n = 1}^{\infty} (3/4)^{n} = \frac{3/4}{1 - 3/4} = 3$ , a finite total — the reference series genuinely converges, so LCT's verdict is trustworthy.

Cell D — a slow logarithm, using the $L = 0$ rule

Worked example Example 4 (Cell D) —

$n = 2 \sum \infty \frac{ln n}{n ^{3}}$ Forecast: $ln n$ grows so slowly that $\frac{l n n}{n ^{3}}$ decays almost like $1/ n^{3}$ . Guess: converges — but we want a clean verdict, so we deliberately choose a comparison that sends $L$ to $0$ .

Pick $b_{n} = \frac{1}{n ^{2}}$ . Why? We compare against a slightly larger but still convergent $p$ -series ( $p = 2 > 1$ ). The extra factor of $n$ in $\frac{1}{n ^{2}}$ versus $\frac{1}{n ^{3}}$ leaves an $n$ in the denominator to swallow the slow $ln n$ , forcing $L = 0$ .
Compute $L$ . $L = lim_{n \to \infty} \frac{( l n n ) / n ^{3}}{1/ n ^{2}} = lim_{n \to \infty} \frac{l n n}{n} .$ This is a classic: a logarithm divided by a positive power of $n$ tends to $0$ (the power of $n$ always wins). So $L = 0$ . Why this step? Because $ln n$ grows slower than $n^{ε}$ for any $ε > 0$ ; here $ε = 1$ so the ratio dies.
Conclude. $L = 0$ and $\sum 1/ n^{2}$ converges. By the $L = 0$ rule of LCT, the series converges. ✔

Verify: the $L = 0$ case only gives convergence when $\sum b_{n}$ converges — and $p = 2 > 1$ guarantees that. Had we carelessly used $b_{n} = 1/ n$ (which diverges) with $L = 0$ , the test would say nothing (see Cell F).

Cell E — $L = \infty$ that still decides

Worked example Example 5 (Cell E) —

$n = 1 \sum \infty \frac{n + 1}{n ^{3/2}}$ Forecast: $\frac{n + 1}{n ^{3/2}} \approx \frac{n}{n ^{3/2}} = \frac{1}{n ^{1/2}}$ , and $p = \frac{1}{2} \leq 1$ diverges. Guess: diverges. We'll show the $L = \infty$ edge case also nails it if we compare against the harmonic series.

Pick $b_{n} = \frac{1}{n}$ (the Harmonic Series, which diverges). Why? We deliberately pick a $b_{n}$ that decays faster than $a_{n}$ , so the ratio blows up — and because $\sum b_{n}$ diverges, the $L = \infty$ rule still applies.
Compute $L$ . $L = lim_{n \to \infty} \frac{( n + 1 ) / n ^{3/2}}{1/ n} = lim_{n \to \infty} \frac{n ( n + 1 )}{n ^{3/2}} = lim_{n \to \infty} \frac{n ^{2} + n}{n ^{3/2}} .$ Divide top and bottom by $n^{3/2}$ : $L = lim_{n \to \infty} (n^{1/2} + n^{- 1/2}) = \infty.$ Why this step? The leading term $n^{2} / n^{3/2} = n^{1/2}$ grows without bound — so $a_{n}$ is eventually much bigger than the harmonic term.
Conclude. $L = \infty$ and $\sum 1/ n$ diverges. By the $L = \infty$ rule, the series diverges. ✔

Verify: since $a_{n}$ is eventually larger than a divergent $1/ n$ , it must diverge too. Cross-check with a friendlier $b_{n} = 1/ n^{1/2}$ : then $L = lim \frac{n + 1}{n} = 1$ , main case, and $\sum 1/ n^{1/2}$ diverges — same verdict, two routes.

Cell F — the degenerate traps

Worked example Example 6 (Cell F) — two ways to break LCT

Trap 6a — the terms don't even go to zero. Consider $n = 1 \sum \infty \frac{n + 1}{2 n + 3}$ .

Forecast: before touching LCT, check the term. Does $a_{n} \to 0$ ?

Compute the term limit. $lim_{n \to \infty} \frac{n + 1}{2 n + 3} = lim_{n \to \infty} \frac{1 + 1/ n}{2 + 3/ n} = \frac{1}{2} \neq = 0.$
Why this step? A convergent series must have $a_{n} \to 0$ (the divergence test / term test). Here $a_{n} \to \frac{1}{2}$ , so the series diverges immediately — no comparison needed. LCT would be a waste of ink.

Verify: $\frac{1}{2} \neq = 0$ , so the necessary condition for convergence fails. Diverges. ✔

Trap 6b — mismatched $L$ . Take the convergent $\sum \frac{1}{n ^{2}}$ but foolishly compare with $b_{n} = \frac{1}{n}$ (harmonic, divergent).

Compute $L$ . $L = lim_{n \to \infty} \frac{1/ n ^{2}}{1/ n} = lim_{n \to \infty} \frac{1}{n} = 0.$
Why this step? $L = 0$ only concludes convergence when $\sum b_{n}$ converges. Here $\sum 1/ n$ diverges, so the $L = 0$ rule gives no information — the test is inconclusive with this $b_{n}$ .
Fix: pick $b_{n} = 1/ n^{2}$ itself (or any convergent $p$ -series with $p > 1$ ); then $L = 1$ , main case, converges cleanly.

Verify: $\sum 1/ n^{2} = π^{2} /6$ converges — the right comparison confirms it; the wrong one just stalls. Lesson: aim for $0 < L < \infty$ , and always match $L = 0$ with convergent $b_{n}$ , $L = \infty$ with divergent $b_{n}$ .

Cell G — a real-world word problem

Worked example Example 7 (Cell G) — total signal energy

A sensor receives, on ping $n$ , an energy of $a_{n} = \frac{4 n + 1}{n ^{3} + 2 n}$ joules. Question: does the total energy $\sum_{n = 1}^{\infty} a_{n}$ over infinitely many pings stay finite (converge) or blow up (diverge)?

Forecast: top $\sim 4 n$ , bottom $\sim n^{3}$ , ratio $\sim 4/ n^{2}$ — like $1/ n^{2}$ , convergent. Guess: total energy is finite.

Pick $b_{n} = \frac{1}{n ^{2}}$ . Why? Strongest power top is $n$ , bottom is $n^{3}$ ; ratio $n / n^{3} = 1/ n^{2}$ .
Compute $L$ . $L = lim_{n \to \infty} \frac{( 4 n + 1 ) / ( n ^{3} + 2 n )}{1/ n ^{2}} = lim_{n \to \infty} \frac{n ^{2} ( 4 n + 1 )}{n ^{3} + 2 n} = lim_{n \to \infty} \frac{4 n ^{3} + n ^{2}}{n ^{3} + 2 n} .$ Divide top and bottom by $n^{3}$ : $L = lim_{n \to \infty} \frac{4 + 1/ n}{1 + 2/ n ^{2}} = 4.$
Conclude. $0 < L = 4 < \infty$ , and $\sum 1/ n^{2}$ converges. The total energy converges — it is finite. ✔

Verify: units check — each $a_{n}$ is in joules, the sum of a convergent joule series is a finite joule total; physically the received energy per ping decays like $1/ n^{2}$ , fast enough to sum to a bounded amount. $L = 4$ is finite and positive, so the verdict is valid.

Cell H — the exam twist (a parameter to tune)

Worked example Example 8 (Cell H) — for which

$p$ does it converge? $n = 1 \sum \infty \frac{n ^{2} + 1}{n ^{p} + n}$ Forecast: top $\sim n^{2}$ , bottom $\sim n^{p}$ (assuming $p > 1$ so $n^{p}$ dominates $n$ ). Ratio $\sim n^{2} / n^{p} = 1/ n^{p - 2}$ — a $p$ -series with exponent $p - 2$ . It converges when $p - 2 > 1$ , i.e. $p > 3$ . Guess: converges iff $p > 3$ .

Pick $b_{n} = \frac{1}{n ^{p - 2}}$ (valid for $p > 1$ so that $n^{p}$ is the dominant denominator term). Why? Keep leading powers: $n^{2} / n^{p} = n^{2 - p} = 1/ n^{p - 2}$ .
Compute $L$ . $L = lim_{n \to \infty} \frac{( n ^{2} + 1 ) / ( n ^{p} + n )}{1/ n ^{p - 2}} = lim_{n \to \infty} \frac{n ^{p - 2} ( n ^{2} + 1 )}{n ^{p} + n} = lim_{n \to \infty} \frac{n ^{p} + n ^{p - 2}}{n ^{p} + n} .$ Divide top and bottom by $n^{p}$ : $L = lim_{n \to \infty} \frac{1 + n ^{- 2}}{1 + n ^{1 - p}} = 1 (for p > 1) .$
Conclude. $L = 1$ always (main case) for $p > 1$ , so $\sum a_{n}$ shares the fate of $\sum 1/ n^{p - 2}$ . That $p$ -series converges $⟺ p - 2 > 1 ⟺ p > 3$ . Boundary check: at $p = 3$ , $b_{n} = 1/ n$ (harmonic) diverges, so the series diverges at $p = 3$ ; for $p < 1$ the denominator is dominated by $n$ instead and the term $\sim n^{2} / n = n$ doesn't even go to $0$ (diverges). So converges exactly when $p > 3$ .

Verify: plug $p = 4$ : $b_{n} = 1/ n^{2}$ , converges — matches. Plug $p = 3$ : $b_{n} = 1/ n$ , diverges — matches boundary. Plug $p = 5$ : $b_{n} = 1/ n^{3}$ , converges — matches. All consistent with "converges $⟺ p > 3$ ." ✔

Recall Which cell is each series?

$\sum \frac{7 n ^{4}}{n ^{7} + 3}$ (poly/poly, converges) ::: Cell A — $b_{n} = 1/ n^{3}$ , $L = 7$ . $\sum \frac{1}{n ^{3} + n}$ ::: Cell B — $b_{n} = 1/ n^{3/2}$ , $p = 3/2 > 1$ , converges. $\sum \frac{2 ^{n} + 1}{5 ^{n}}$ ::: Cell C — geometric, $r = 2/5$ , converges. $\sum \frac{l n n}{n ^{4}}$ ::: Cell D — $L = 0$ vs convergent $1/ n^{3}$ , converges. $\sum \frac{n ^{2}}{n ^{3/2}}$ ::: Cell F — terms $\to \infty$ , diverges instantly.

See also: Ratio Test and Integral Test for series where no clean $p$ -series or geometric skeleton exists, and the Squeeze Theorem which powers the $ε = L /2$ sandwich behind LCT.

Recall

Recall Self-test

In Cell C, why can we ignore $n^{2}$ next to $3^{n}$ ? ::: Exponential growth (base $> 1$ ) eventually dominates any polynomial, so $n^{2} / 3^{n} \to 0$ . In Cell E, why does $L = \infty$ still give a verdict? ::: Because $\sum b_{n}$ (harmonic) diverges, and the $L = \infty$ rule then forces $\sum a_{n}$ to diverge. In Cell H, at the boundary $p = 3$ does the series converge? ::: No — $b_{n} = 1/ n$ is harmonic, diverges. What breaks in Cell F trap 6a? ::: Terms tend to $1/2 \neq = 0$ , so the divergence test kills it before LCT.