Convergence tests for improper integrals — comparison: Question bank

Read each prompt, commit to an answer out loud, then reveal. If your justification differs from mine, that gap is the misconception to fix.

True or false — justify

A bound

f < g

with

\int g

divergent proves

\int f

diverges.

False. Being below an infinite area can still be finite (e.g.

1/ x^{2} < 1/ x

, yet

\int 1/ x^{2}

converges). A useful lower bound must sit above a divergent

g

0 \leq f \leq g

and

\int f

diverges, then

\int g

diverges.

True. If the smaller area is already infinite, the larger area (which is

\geq

it) cannot be finite — this is the divergence half of the DCT.

The DCT works for any two functions as long as

f \leq g

False. It also needs

f \geq 0

(equivalently

0 \leq f \leq g

); the proof relies on

F (t) = \int_{a}^{t} f

being increasing, which fails if

f

changes sign.

\int_{a}^{\infty} ∣ f ∣ d x

converges, then

\int_{a}^{\infty} f d x

converges.

True. This is absolute convergence; controlling

∣ f ∣

by a convergent

g

lets us handle sign-changing

f

— see Absolute vs conditional convergence.

\int_{a}^{\infty} f d x

converges, then

\int_{a}^{\infty} ∣ f ∣ d x

must converge too.

False. Convergence can be conditional: cancellation of positive and negative area can make

\int f

finite while

\int ∣ f ∣

diverges.

\int_{1}^{\infty} x^{- p} d x

converges for

p = 1

False. At exactly

p = 1

the antiderivative is

ln t \to \infty

, so it diverges. The boundary belongs to the divergent side; convergence needs

p > 1

strictly.

For

\int_{0}^{1} x^{- p} d x

(a type-2 improper integral at

0

), convergence also needs

p > 1

False. Near

0

the rule flips:

\int_{0}^{1} x^{- p} d x

converges iff

p < 1

. Small powers are integrable near

0

, large powers near

\infty

— see Improper integrals — infinite discontinuities (type 2).

lim_{x \to \infty} f (x) = 0

, then

\int_{1}^{\infty} f d x

converges.

False. The integrand vanishing is necessary-feeling but not sufficient:

1/ x \to 0

yet

\int_{1}^{\infty} 1/ x d x

diverges. Speed of decay is what matters.

The Limit Comparison Test requires

f

and

g

to be eventually positive.

True. LCT is stated for

f, g > 0

for large

x

; the ratio-sandwich argument

\frac{L}{2} g \leq f \leq 2 Lg

only makes sense with positive functions.

L = lim f / g = 1

, then

\int f

and

\int g

literally have equal value.

False.

0 < L < \infty

guarantees the same fate (both converge or both diverge), never equal numerical values — only that their tails scale proportionally.

Spot the error

\frac{1}{ln x} < \frac{1}{x}

for large

x

, so since

\int 1/ x

diverges,

\int 1/ ln x

diverges."

The inequality is backwards: since

ln x < x

, we get

\frac{1}{l n x} > \frac{1}{x}

. The corrected bound (bigger than a divergent integral) is exactly what forces divergence.

\frac{2 + sin x}{x ^{2}} \leq \frac{2}{x ^{2}}

, and

\int 2/ x^{2}

converges, so the integral converges."

The bound is false where

sin x > 0

. The safe envelope is

2 + sin x \leq 3

, giving

\leq 3/ x^{2}

; the conclusion survives but only with the correct constant.

"I want convergence, so I found

g

with

f \geq g

and

\int g

convergent — done."

Wrong direction: bounding

f

below a convergent

g

tells you nothing about

f

. To prove convergence you must bound

f

above by a convergent

g

"LCT gave

L = 0

, and

\int g

diverges, so

\int f

diverges."

With

L = 0

you only get the one-way link "

\int g

converges

\Rightarrow \int f

converges." Divergence of

g

says nothing;

f

decays faster than

g

, so it may well converge.

f \leq g

pointwise everywhere except near

x = 3

, but that's fine, comparison still applies globally."

Comparison needs the inequality to hold for all $x \geq a$ (or eventually, adjusting

a

). A single interval where

f > g

can add extra area, so you must restrict to where the bound truly holds.

\int_{1}^{\infty} \frac{s i n x}{x ^{2}} d x

: since

sin x

isn't positive, DCT can't say anything, so it diverges."

The premise is right (DCT needs

f \geq 0

) but the leap is wrong. Use

\frac{s i n x}{x ^{2}} \leq \frac{1}{x ^{2}}

; absolute comparison gives convergence.

"Comparing

\frac{1}{x + x}

\frac{1}{x}

by LCT: dominant term is

x

The dominant term as

x \to \infty

x

(it grows faster than

x

), so the correct benchmark is

1/ x

, giving

L = 1

and divergence — not

1/ x

Why questions

Why must

f \geq 0

for the Direct Comparison Test but not obviously for the Limit Comparison Test?

DCT's proof uses that

F (t) = \int_{a}^{t} f

is increasing (needs

f \geq 0

); LCT also secretly needs positivity so the ratio bound

\frac{L}{2} g \leq f \leq 2 Lg

is a valid two-sided squeeze — both rely on nonnegativity via Monotone Convergence Theorem.

Why do we define

\int_{a}^{\infty} f = lim_{t \to \infty} \int_{a}^{t} f

instead of just "integrating to infinity"?

\infty

is not a number you can substitute into an antiderivative; you integrate over a legitimate finite window

[a, t]

and ask whether the accumulated area settles as

t

grows.

Why is

1/ x^{2}

the go-to convergent benchmark and

1/ x

the go-to divergent one?

They straddle the

p = 1

threshold:

p = 2 > 1

converges,

p = 1

diverges. Any tail behaving "faster than

1/ x

" leans convergent, "like

1/ x

or slower" leans divergent — see The p-integral and p-series.

Why does the comparison test for integrals mirror the comparison test for series?

Both accumulate nonnegative pieces and rely on "increasing + bounded

\Rightarrow

converges"; the Integral Test for series makes the correspondence exact for monotone terms.

Why does bounding a wiggling numerator like

2 + sin x

by a constant work?

Because

sin x

is trapped in

[- 1, 1]

, so

2 + sin x \in [1, 3]

; replacing the oscillation by its worst-case constant

3

gives a clean comparison without losing correctness.

Why can't "

f (x) \to 0

" alone guarantee convergence, when it is required for series?

Even for series

a_{n} \to 0

is only necessary, not sufficient (

\sum 1/ n

diverges); integrals inherit the same lesson — the rate of decay decides, not the mere fact of decay.

Why does LCT free us from choosing the correct inequality direction?

It converts the problem into a single limit

L = lim f / g

; once

0 < L < \infty

, the two-sided squeeze automatically supplies both DCT inequalities, so you never have to guess which way the bound points.

Edge cases

What happens in LCT when

L = \infty

f

dominates

g

: you only get "

\int f

converges

\Rightarrow \int g

converges" (equivalently "

\int g

diverges

\Rightarrow \int f

diverges"). It's a one-way tool, useful only if

g

diverges.

What happens in LCT when

L = 0

f

is negligible beside

g

: you only get "

\int g

converges

\Rightarrow \int f

converges." If

g

diverges you learn nothing about

f

An integral is improper in two ways at once (e.g.

\int_{0}^{\infty} x^{- 1} d x

). How do you test it?

Split at a convenient interior point, say

\int_{0}^{1} + \int_{1}^{\infty}

; the whole converges only if both pieces converge. Here

\int_{0}^{1}

diverges (

p = 1

0

) and

\int_{1}^{\infty}

diverges, so it diverges either way.

Does

\int_{1}^{\infty} x^{- p} d x

converge at

p

slightly above

1

, like

p = 1.0001

Yes. The criterion is

p > 1

strictly, with no gap: any

p > 1

, however close, gives

t^{1 - p} \to 0

and a finite value. Convergence is razor-thin but genuine.

f \leq g

only holds for

x \geq M

(not from

x = a

), can we still conclude convergence?

Yes — convergence depends only on the tail. Split

\int_{a}^{\infty} = \int_{a}^{M} + \int_{M}^{\infty}

; the finite first piece is harmless, and DCT applies to

\int_{M}^{\infty}

where the bound holds.

What if

f \geq 0

but

f

is unbounded on the finite part while decaying at infinity?

Then it's improper of both types; test each singular point separately. Convergence at

\infty

(via comparison to

x^{- p}

p > 1

) and integrability at the blow-up point are independent conditions — both must hold.

Can two integrals both diverge yet have a finite ratio limit

L

Yes, that's the whole point of LCT:

\int 1/ x

and

\int 1/ (x + x)

both diverge with

L = 1

. Same fate does not mean finite; it means they diverge together.

Connections

The p-integral and p-series
Comparison test for infinite series
Absolute vs conditional convergence
Monotone Convergence Theorem
Integral Test for series
Improper integrals — infinite discontinuities (type 2)