Ratio test — proof, limitations: Question bank

Before the questions, one shared mental picture — and note it introduces the letter $r$ , which we must earn:

Recall How the picture becomes the formal

$ε$ -proof "Settles below the $r$ -line" is exactly the $ε$ -statement: with $ε = r - L > 0$ , the limit gives an $N$ so that $n \geq N \Rightarrow ∣ a_{n + 1} / a_{n} ∣ < L + ε = r$ . Chaining that inequality $k$ times yields $∣ a_{N + k} ∣ < r^{k} ∣ a_{N} ∣$ — the geometric wall. Then $\sum_{k} r^{k} ∣ a_{N} ∣ = ∣ a_{N} ∣/ (1 - r) < \infty$ because $r < 1$ , and the comparison test transfers that finiteness to $\sum ∣ a_{N + k} ∣$ . So the coarse phrase "the wall shrinks" is the precise reason bounding by $r^{k}$ produces a convergent comparison series.

True or false — justify

TF1. If

lim_{n \to \infty} ∣ a_{n + 1} / a_{n} ∣ = 1

then

\sum a_{n}

diverges.

False.

L = 1

is the inconclusive case;

\sum 1/ n^{2}

has

L = 1

and converges while

\sum 1/ n

has

L = 1

and diverges — same

L

, opposite fate.

TF2. If

\sum a_{n}

converges then its ratio limit

L

must be less than

1

False. Convergence does not force

L < 1

;

\sum 1/ n^{2}

converges yet its

L = 1

. Only

L < 1

implies convergence, not the reverse.

TF3. The ratio test can prove conditional convergence of the alternating harmonic series

\sum (- 1)^{n} / n

False. The ratio uses absolute values and here

L = 1

, so the test is silent; conditional convergence needs the alternating series test instead.

TF4. If the ratio test gives

L > 1

, the root test could still return

L = 1

on the same series.

False. When both limits exist they agree; if the ratio limit is

> 1

the root limit is the same value, so the root test also diverges (never

1

TF5. A series of strictly positive terms with

L < 1

converges "absolutely".

True, and trivially so — for positive terms

∣ a_{n} ∣ = a_{n}

, so absolute convergence and plain convergence are the same statement.

TF6. If

∣ a_{n + 1} / a_{n} ∣ < 1

for every

n

, the series must converge.

False. The ratio being below

1

isn't enough; it must stay below a fixed

r < 1

. Harmonic

\sum 1/ n

has

n / (n + 1) < 1

always yet diverges because those ratios creep toward

1

TF7. If the shrink-ratios oscillate and

lim_{n \to \infty} ∣ a_{n + 1} / a_{n} ∣

does not exist, the ratio test simply reports "inconclusive".

True in the basic (limit) form. No limit means the theorem's hypothesis fails — but the sharper

lim sup

ratio test can still act: if

lim sup ∣ a_{n + 1} / a_{n} ∣ < 1

it converges, and if

lim inf ∣ a_{n + 1} / a_{n} ∣ > 1

it diverges (see EC7).

TF8.

L = \infty

and

L > 1

lead to the same conclusion.

True. Both mean terms eventually grow, so

a_{n} \neq \to 0

and the series diverges by the term test.

Spot the error

SE1. "For

\sum (- 1)^{n} / 2^{n}

, the ratio is

a_{n + 1} / a_{n} = - 1/2

, and since

- 1/2 < 1

the series converges." — find the flaw.

The verdict is right but the reasoning is broken: you must use

∣ a_{n + 1} / a_{n} ∣ = 1/2

. A raw ratio of

- 1/2

is not what the theorem compares against, and negative ratios can't bound a geometric wall.

SE2. "

\sum 1/ n^{3}

: since

a_{n + 1} / a_{n} = n^{3} / (n + 1)^{3} \to 1

, the ratio test says it diverges." — find the flaw.

L = 1

means no information, not divergence. In fact

\sum 1/ n^{3}

converges (a p-series with

p = 3 > 1

); the ratio test is just the wrong tool here.

SE3. "The terms of

\sum 1/ n

go to zero, so the ratio test's

L < 1

branch applies and it converges." — find the flaw.

Two errors:

a_{n} \to 0

is not the ratio-test condition, and here

L = 1

(not

< 1

). Terms going to zero is necessary for convergence but never sufficient.

SE4. "In the

L < 1

proof we pick any

r

with

L < r < 1

; if

L = 0

there is no such

r

, so the proof fails for

L = 0

." — find the flaw.

There are such

r

when

L = 0

— take

r = 1/2

, any number strictly between

0

and

1

works. The gap

(L, 1)

is nonempty whenever

L < 1

SE5. "Because

∣ a_{N + k} ∣ < r^{k} ∣ a_{N} ∣

, the whole series equals

∣ a_{N} ∣/ (1 - r)

." — find the flaw.

That value is an upper bound on the tail, not the sum. The real sum is smaller, and we also have finitely many terms before

N

; comparison only guarantees finiteness, not a value.

SE6. "

\sum n! /1 0^{n}

diverges because

n!

is huge, so its ratio limit is some big constant like

10

." — find the flaw.

The ratio is

(n + 1) /10

, which grows without bound, so

L = \infty

, not a fixed constant. Factorials eventually beat any fixed exponential — the ratio doesn't level off.

SE7. "Applying the ratio test to

\sum 2^{n}

gives

a_{n + 1} / a_{n} = 2 > 1

, so it diverges — but only because it's not a geometric series." — find the flaw.

\sum 2^{n}

is geometric with ratio

2

; the ratio test just recovers the geometric rule. The parenthetical claim is false, not a caveat.

Why questions

WHY1. Why does the proof compare against a geometric series specifically, not a p-series?

Because a fixed shrink-factor

r

is exactly what defines a geometric series, and we know precisely when it converges (

∣ r ∣ < 1

). The ratio test measures that very factor, so geometric is the natural yardstick.

WHY2. Why must the trapping

r

be a fixed number rather than "the ratio at each step"?

A per-step ratio can drift up to

1

(like harmonic's

n / (n + 1)

) and never let the geometric wall shrink. Only a single fixed

r < 1

forces

r^{k} \to 0

and hence a finite total.

WHY3. Why does

L > 1

guarantee divergence but

L < 1

does not guarantee the terms are monotonically shrinking?

L > 1

forces

∣ a_{n + 1} ∣ > ∣ a_{n} ∣

eventually, so

a_{n} \neq \to 0

— a hard failure. But

L < 1

only says the long-run ratio is small; early terms may bounce around before the geometric domination kicks in past

N

WHY4. Why is the ratio test "blind" to every power-law series

\sum 1/ n^{p}

For any

p

a_{n + 1} / a_{n} = (n / (n + 1))^{p} \to 1

, landing squarely in the

L = 1

dead zone. The test can't see the exponent

p

, which is exactly what decides convergence there.

WHY5. Why does the ratio test shine on factorials and exponentials?

\frac{a _{n + 1}}{a _{n}}

the factorials telescope (

n! / (n + 1)! = 1/ (n + 1)

) and exponentials collapse to a constant (

r^{n + 1} / r^{n} = r

), leaving a clean limit instead of a stubborn

1

WHY6. Why is the ratio test the tool of choice for finding a power series' radius of convergence?

Setting

L = lim_{n \to \infty} ∣ a_{n + 1} / a_{n} ∣ \cdot ∣ x ∣ < 1

solves directly for the range of

∣ x ∣

; the boundary

L = 1

marks the radius, exactly where geometric domination breaks.

WHY7. Why can Raabe's test sometimes decide cases where the ratio test returns

L = 1

Raabe's test zooms into the rate at which the ratio approaches

1

(via

n (1 - a_{n + 1} / a_{n})

), a finer measurement that distinguishes

1/ n

from

1/ n^{2}

where the coarse ratio limit cannot.

WHY8. Why does the term-by-term comparison step actually finish the convergence proof?

Once

∣ a_{N + k} ∣ \leq r^{k} ∣ a_{N} ∣

and

\sum r^{k}

converges, comparison says a series dominated by a convergent positive series is itself (absolutely) convergent — that theorem does the closing.

Edge cases

EC1. What does the ratio test say about a series with a zero term in the middle, like

a_{5} = 0

The basic ratio

a_{n + 1} / a_{n}

is undefined at that step (division by zero). You only need the terms to be eventually nonzero; a stray zero early on is fine since the limit only cares about the tail.

EC2. What if

a_{n} = 0

for all

n

beyond some point?

Then the series is a finite sum — trivially convergent — and the ratio test isn't needed (indeed

a_{n + 1} / a_{n}

becomes

0/0

). Recognise the degenerate case rather than force the test.

EC3. Ratio test verdict when

L = 0

Converges absolutely, and very strongly —

L = 0

means terms crush faster than any geometric series, so it's the "most convergent" outcome, not a special exception.

EC4. Two series both hit

L = 1

\sum 1/ n

and

\sum 1/ n^{2}

. What single sharper fact separates them?

The exponent

p

in the p-series rule:

p = 1

diverges,

p = 2 > 1

converges. The ratio test can't read

p

; the integral/p-test can.

EC5. At the endpoints

∣ x ∣ = 1

of a power series where

L = 1

, what must you do?

Test each endpoint by hand as its own series; the ratio test has already given up there. E.g. for

\sum x^{n} / n

x = 1

diverges (harmonic) but

x = - 1

converges (alternating harmonic).

EC6. If a series has

L < 1

but individual ratios exceed

1

for the first hundred terms, does it still converge?

Yes. Convergence depends only on the tail past some

N

; a finite block of growing terms adds a finite amount and cannot spoil convergence once the ratio settles below a fixed

r < 1

EC7. The plain limit

lim_{n \to \infty} ∣ a_{n + 1} / a_{n} ∣

doesn't exist — is the ratio test truly useless?

Not always. The

lim sup

variant still decides the convergent case: if

lim sup ∣ a_{n + 1} / a_{n} ∣ < 1

you can still trap the tail under a fixed

r < 1

from some

N

on, so it converges; and if

lim inf ∣ a_{n + 1} / a_{n} ∣ > 1

the terms eventually grow, so it diverges. Only when the limsup straddles

1

is it genuinely inconclusive.

Connections

Ratio test — proof, limitations — the parent this bank stress-tests.
Geometric Series — the yardstick every $L$ comparison uses.
Comparison Test — closes the $L < 1$ convergence case.
Term Test (nth-term divergence) — kills the $L > 1$ case.
Root Test — the sibling that agrees when both limits exist.
p-Series and Integral Test · Raabe's Test — sharper rulers for the $L = 1$ dead zone.
Radius of Convergence — the ratio test's headline application.