Limit comparison test: Visual walkthrough — Maths deep dive

This page rebuilds the Limit Comparison Test (LCT) with nothing assumed. Every symbol is earned before it is used, and every step has a picture. If you have never seen a series before, start at Step 1 — we build the whole idea from a pile of dots.

Step 0 — What is a "series" and what do the symbols mean?

Before any formula, let's name the objects.

$a_{1}, a_{2}, a_{3}, \dots$ is a sequence: an ordered list of numbers, one for each counting position $n = 1, 2, 3, \dots$ . Think of $a_{n}$ as "the $n$ -th bar's height."
$\sum a_{n}$ (read "sum of the $a_{n}$ ") means: add all those heights, $a_{1} + a_{2} + a_{3} + \dots$ , forever.
Positive-term means $a_{n} > 0$ and $b_{n} > 0$ for all large $n$ : from some position onward every bar points up, none points down, none is zero. (A finite handful of early terms may misbehave — they can't change convergence, which is decided by the tail.) We will lean on $b_{n} > 0$ hard, so it is a standing assumption from here on.

In the figure the red curve is the running total. On the left it flattens toward a ceiling (converges); on the right it keeps rising (diverges). The individual bars can both shrink toward zero — so "bars shrink to zero" alone cannot tell the two apart. That is exactly the trap the LCT avoids.

Step 1 — The one question the test answers

WHAT. We have $\sum a_{n}$ whose fate is unknown, and $\sum b_{n}$ whose fate we already know (a p-Series or a Geometric Series). We ask: do $a_{n}$ and $b_{n}$ shrink at the same speed? Both are positive (Step 0).

WHY. For large $n$ the fate of a series is decided by how fast its terms decay, not by their exact values. So the right measuring instrument is the ratio $a_{n} / b_{n}$ — it strips away the sizes and keeps only the relative decay speed. Dividing is only legal because $b_{n} > 0$ (never zero).

PICTURE. Two shrinking bar-sequences drawn on the same axis, both positive. Their heights differ, but their shape of decay is what we compare.

If $L$ is a finite positive number, the sequences are proportional in the long run: $a_{n} \approx L \cdot b_{n}$ . That single fact is the whole engine.

Step 2 — Enter the main case, and turn "the limit is $L$ " into a window

WHAT. From here through Step 5 we work in the main case: $0 < L < \infty$ , i.e. $L$ is a finite, strictly positive number. Under that assumption we convert " $a_{n} / b_{n} \to L$ " into a concrete band that traps the ratio for all large $n$ . (The edge cases $L = 0$ and $L = \infty$ get their own treatment in Step 6.)

WHY. A limit is a promise: pick any tiny tolerance, and eventually the ratio stays within that tolerance of $L$ . We cash in that promise with a specific tolerance we choose on purpose. We choose $ε = L /2$ — and this choice only makes sense because $L > 0$ (the main-case assumption). If $L$ were $0$ , then $L /2$ would be $0$ and there would be no room for a band; that is precisely why $L = 0$ is a separate case.

So, still assuming $0 < L < \infty$ , pick the tolerance $ε = L /2$ (half of $L$ ). By the meaning of a limit, there is a cutoff position $N$ so that for every $n > N$ : $how far the ratio is from L \frac{a _{n}}{b _{n}} - L < our chosen tolerance \frac{L}{2} .$

PICTURE. A horizontal line at height $L$ , with a shaded band of half-width $L /2$ around it. Past the cutoff $N$ (dashed vertical line), every ratio dot lands inside the band.

Step 3 — Unpack the absolute value into two plain inequalities

WHAT. $∣ x ∣ < c$ is shorthand for " $x$ is squeezed between $- c$ and $+ c$ ." We apply that here (still in the main case $0 < L < \infty$ ).

WHY. Absolute value hides two facts at once (an upper bound and a lower bound). We need them separately so we can use each one for a different conclusion later.

$\frac{a _{n}}{b _{n}} - L < \frac{L}{2} ⟹ - \frac{L}{2} < \frac{a _{n}}{b _{n}} - L < \frac{L}{2} .$ Add $L$ to all three parts: $lower fence \frac{L}{2} < \frac{a _{n}}{b _{n}} < upper fence \frac{3 L}{2} (n > N) .$

$\frac{L}{2}$ = the floor: the ratio never drops below this.
$\frac{3 L}{2}$ = the ceiling: the ratio never rises above this.

PICTURE. The same band as before, now with its floor and ceiling labelled $L /2$ and $3 L /2$ — a corridor the ratio is trapped inside.

Step 4 — Clear the fraction to expose a direct comparison

WHAT. Multiply every part of the double inequality by $b_{n}$ .

WHY. Right now we have a statement about the ratio. To use the Direct Comparison Test we need statements about $a_{n}$ itself sitting between multiples of $b_{n}$ . Multiplying by $b_{n}$ does exactly that.

Crucial: $b_{n} > 0$ (our standing assumption from Step 0), so multiplying does not flip any inequality sign. If $b_{n}$ could be negative or zero, the whole argument would collapse — this is why positivity is non-negotiable.

$lower wall \frac{L}{2} b_{n} < our unknown term a_{n} < upper wall \frac{3 L}{2} b_{n} (n > N) .$

PICTURE. For each $n$ past the cutoff, the bar $a_{n}$ (red) is boxed between a short bar $\frac{L}{2} b_{n}$ and a tall bar $\frac{3 L}{2} b_{n}$ — both just scaled copies of $b_{n}$ .

Step 5 — Feed each wall into the Direct Comparison Test

WHAT. Use the two walls separately, depending on what we know about $\sum b_{n}$ .

WHY. We invoke the Direct Comparison Test here, so let's recall exactly what it says before using it.

Our sandwich hands us both a below-fact ( $a_{n} < \frac{3 L}{2} b_{n}$ ) and an above-fact ( $a_{n} > \frac{L}{2} b_{n}$ ), so exactly one branch fires depending on $\sum b_{n}$ .

Case A — $\sum b_{n}$ converges. Then $\sum \frac{3 L}{2} b_{n}$ converges too (multiplying every term by the constant $\frac{3 L}{2}$ can't create an infinite total). Since $a_{n} < \frac{3 L}{2} b_{n}$ , the term $a_{n}$ is capped by a convergent series, so $\sum a_{n} converges.$

Case B — $\sum b_{n}$ diverges. Then $\sum \frac{L}{2} b_{n}$ diverges (shrinking every term by the constant $\frac{L}{2}$ can't rescue an infinite total). Since $a_{n} > \frac{L}{2} b_{n}$ , the term $a_{n}$ dominates a divergent series, so $\sum a_{n} diverges.$

Either way, $\sum a_{n}$ and $\sum b_{n}$ live or die together. $■$

PICTURE. Two running-total curves drawn together: $a_{n}$ 's total (red) is pinned between the totals of $\frac{L}{2} b_{n}$ and $\frac{3 L}{2} b_{n}$ . If the outer pair flattens, red flattens; if the inner one climbs, red climbs.

Step 6 — The edge cases $L = 0$ and $L = \infty$

WHAT. When the ratio limit is $0$ or $\infty$ , only one wall of the sandwich survives — so the test gives one-directional information.

WHY. If $L = 0$ , the ratio $a_{n} / b_{n}$ shrinks past any bound, so eventually $a_{n} < b_{n}$ : the upper wall exists but the lower wall has collapsed to zero. If $L = \infty$ , the ratio explodes, so eventually $a_{n} > b_{n}$ : only the lower wall survives.

$L = 0$ and $\sum b_{n}$ converges → $a_{n}$ is smaller than a convergent series → $\sum a_{n}$ converges.
$L = \infty$ and $\sum b_{n}$ diverges → $a_{n}$ is bigger than a divergent series → $\sum a_{n}$ diverges.

The mismatched pairs ( $L = 0$ with $\sum b_{n}$ divergent; $L = \infty$ with $\sum b_{n}$ convergent) give no information — the surviving wall points the wrong way.

PICTURE. Left panel: $L = 0$ , the $a_{n}$ bar hugs the floor beneath $b_{n}$ (only an upper cap). Right panel: $L = \infty$ , the $a_{n}$ bar towers above $b_{n}$ (only a lower floor).

Step 7 — The degenerate case: the limit does not exist

WHAT. Every step above quietly used one fact: the ratio $a_{n} / b_{n}$ settles down to some value $L$ . If it does not settle — if it keeps oscillating between different heights forever — then there is no $L$ , and the whole machine has nothing to grab onto.

WHY. The proof turns "the limit is $L$ " into a band around $L$ (Step 2). With no $L$ , there is no centre for the band, so no cutoff $N$ past which the ratio is trapped. The sandwich of Step 4 can never be built.

What to do instead. Choose a different $b_{n}$ so that the new ratio does converge, or abandon LCT for the Ratio Test, Integral Test, or Direct Comparison Test.

PICTURE. Ratio dots that keep bouncing between a high level and a low level, never entering any fixed band — the cutoff line can be drawn anywhere and dots still escape.

A quick numeric sanity check of the sandwich

The one-picture summary

Everything above collapses into a single image: provided the ratio has a limit, the ratio $a_{n} / b_{n}$ enters a positive band $(L /2, 3 L /2)$ , which becomes the term-sandwich $\frac{L}{2} b_{n} < a_{n} < \frac{3 L}{2} b_{n}$ , which hands each wall to Direct Comparison — so the two series share one fate.

Recall Feynman retelling — say it like a story

Imagine two lines of shrinking blocks, the $a$ -line (mystery) and the $b$ -line (known) — both blocks are always positive from some point on, that's a rule I never break. I only care whether they shrink at the same speed, so I divide one by the other (safe, since the $b$ -block is never zero) and watch what that ratio settles down to — call it $L$ . First I must check it actually settles: if the ratio just bounces around forever, there is no $L$ and this whole method is off the table. In the main case $L$ is a normal positive number, so the blocks are basically proportional forever. I draw a band around $L$ that's half as wide as $L$ is tall — I'm only allowed to do that because $L$ is bigger than zero, which is why $L = 0$ is its own separate case. That width keeps the band's floor above zero. The limit promises that after some point every ratio falls inside that band. I multiply the band by the (positive!) $b$ -block height, and now each mystery block $a_{n}$ is boxed between a short copy and a tall copy of the known block. Direct Comparison then finishes it: if the known series adds up to something finite, the tall copies do too, so the mystery — being even smaller — must also add up. If the known series blows up, the short copies blow up, so the mystery — being even bigger — blows up too. Same fate, proven. The three places this could break: negative or zero blocks (dividing/multiplying goes wrong), a ratio with no limit (nothing to centre the band on), and the two lopsided cases $L = 0$ or $L = \infty$ , where one wall of the box vanishes and you only get half the story.

Recall Self-test

Why must $b_{n} > 0$ ? ::: So dividing is defined and multiplying the ratio inequality by $b_{n}$ preserves direction; Direct Comparison also needs positive terms. Why pick $ε = L /2$ , and why does it need $L > 0$ ? ::: Any $0 < ε < L$ keeps the floor $L - ε$ positive; $L /2$ is the tidy choice giving floor $L /2 > 0$ — and this is only possible in the main case $L > 0$ . In Case A, which wall is used and why? ::: The upper wall $a_{n} < \frac{3 L}{2} b_{n}$ , capping $a_{n}$ under a convergent series. If $L = 0$ but $\sum b_{n}$ diverges, what do you conclude? ::: Nothing — the surviving upper wall points the wrong way; the test is inconclusive. What if $lim a_{n} / b_{n}$ does not exist? ::: The test does not apply — pick a different $b_{n}$ or use another test.

Parent: Limit comparison test · See also: Direct Comparison Test, Squeeze Theorem, p-Series, Geometric Series, Harmonic Series