Integral test — proof, p-series: Question bank

Tag: #calculus3 #series/convergence

This is a concept gym for the integral test and the p-series. Every item below targets a place where intuition lies to you. Cover the right side, force a full sentence of reasoning, then reveal. Bare "yes/no" answers earn nothing — the reasoning is the point.

Before we start, three words we lean on constantly, defined from zero:

Keep these two pictures in front of you the whole time. Figure 1 shows why the trap works when $f$ decreases; Figure 2 shows why it breaks when $f$ increases. Nearly every item below is really one of these two pictures in words.

Figure — Integral test — proof, p-series

True or false — justify

Every answer must say why, not just T/F.

TF1. A p-series with

p = 0.9999

converges.

False. Convergence needs

p > 1

strictly;

0.9999 < 1

\int_{1}^{\infty} x^{- 0.9999} d x

still blows up. The cutoff is razor-sharp, and

0.9999

is on the diverging side.

TF2. If

\int_{1}^{\infty} f d x = 5

, then

\sum_{n = 1}^{\infty} f (n) = 5

too.

False. The test only guarantees the sum and integral share a fate (both finite), not a value. Look at Figure 1: the red left-endpoint rectangles sit strictly above the curve for a decreasing

f

, so

\sum f (n) > \int_{1}^{\infty} f

in general (e.g.

\sum 1/ n^{2} = π^{2} /6 \approx 1.645

vs integral

= 1

TF3. The harmonic series diverges even though its terms

1/ n \to 0

True. Terms shrinking to

0

is necessary for convergence but not sufficient; the shrinking is too slow, and

\int_{1}^{\infty} d x / x = ln b \to \infty

confirms divergence.

TF4. If

a_{n} \to 0

, then by the integral test

\sum a_{n}

converges.

False. That is not what the integral test says, and it's the classic error.

a_{n} \to 0

only lets you avoid instant divergence; you still must run a real test. Harmonic series is the counterexample.

TF5. The integral test can be used on

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

False. The test demands

f

positive; alternating terms have mixed signs, so the "areas can't cancel" assumption is broken. Use an alternating-series test instead.

TF6. If

f

is positive and continuous but increasing, the integral test still applies.

False. Look at Figure 2: when

f

increases, on

[n, n + 1]

the left endpoint gives the smallest value, so the left-endpoint rectangle now lies below the curve, giving

f (n) \leq \int_{n}^{n + 1} f

— the exact reverse of the decreasing case

\int_{n}^{n + 1} f \leq f (n)

. Both trap-inequalities flip, so

S_{N}

is no longer squeezed the right way and the test's hypothesis is genuinely violated. (Separately, an increasing positive

f

has

a_{n} \neq \to 0

, so the series diverges by the n-th term test for divergence anyway.)

TF7.

\sum_{n = 1}^{\infty} \frac{1}{n ^{p}}

diverges for every

p \leq 1

, including negative

p

True. For

p \leq 1

the integral

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

is infinite. For

p \leq 0

the terms don't even shrink to

0

, so it diverges instantly by the n-th term test — divergence for all

p \leq 1

TF8. The Riemann zeta function

ζ (p) = \sum 1/ n^{p}

is defined for

p = 1

False.

ζ

is defined by this sum only for

p > 1

, exactly where the p-series converges. At

p = 1

the sum is the divergent harmonic series.

TF9. If

\sum f (n)

converges, then

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

must converge too.

True — provided $f$ is positive, continuous and decreasing (the integral-test hypotheses; without them the equivalence is not guaranteed). Under those conditions the equivalence runs both ways: convergence of the sum forces convergence of the integral via the lower-sum inequality

\int_{1}^{N + 1} f \leq S_{N}

, whose left side stays bounded. They live or die together.

Spot the error

Each line states a flawed argument; the reveal names the exact broken step.

SE1. "

\sum \frac{1}{n l n n}

converges because

\frac{1}{n l n n}

shrinks faster than

\frac{1}{n}

The error is trusting "shrinks faster" without a test. The substitution

u = ln x

gives

\int_{2}^{\infty} \frac{d x}{x l n x} = \int_{l n 2}^{\infty} \frac{d u}{u} = \infty

, so it diverges — the extra

ln n

isn't enough to save it.

SE2. "

f (x) = \frac{s i n ^{2} x}{x ^{2}}

is positive and continuous, so I'll apply the integral test."

Missing hypothesis:

f

is not decreasing (it oscillates because of

sin^{2} x

). The integral test requires eventually-decreasing

f

; here you'd need a comparison with

1/ x^{2}

instead.

SE3. "For

\sum 1/ n^{2}

, since

\int_{1}^{\infty} x^{- 2} d x = 1

, the sum equals

1

The error equates sum and integral. The test only bounds:

1 = \int_{1}^{\infty} x^{- 2} d x < \sum 1/ n^{2} = π^{2} /6 \approx 1.645

. Sharing convergence

\neq =

sharing value.

SE4. "

\int_{1}^{\infty} x^{- p} d x = \frac{x ^{1 - p}}{1 - p}_{1}^{\infty}

, and plugging in

p = 1

gives

\frac{1}{0}

, undefined but finite-ish, so harmonic is borderline convergent."

The error is using the

p \neq = 1

antiderivative at

p = 1

. When

p = 1

the antiderivative is

ln x

, not

x^{1 - p} / (1 - p)

; and

ln b \to \infty

, so harmonic diverges outright — not "borderline convergent."

SE5. "The rectangles equal the curve's area, so the integral test is exact."

Rectangles of a decreasing

f

never equal the curve's area — left-endpoint rectangles overshoot, right-endpoint ones undershoot (see Figure 1). The gap is exactly what the trapping inequality

\int_{1}^{N + 1} f \leq S_{N} \leq f (1) + \int_{1}^{N} f

measures.

SE6. "

\sum_{n = 1}^{\infty} \frac{1}{n l n n}

— apply the test from

n = 1

n = 1

ln 1 = 0

f (1) = \frac{1}{1 \cdot 0}

is undefined. Start the test from

n = 2

; the test is happy to begin at any

n = N

since finitely many terms never change convergence.

Why questions

WHY1. Why must

f

be decreasing, not just positive and continuous?

Decreasing guarantees the left-endpoint height

f (n)

is the max on

[n, n + 1]

and the right-endpoint height

f (n + 1)

is the min (Figure 1). That gives both one-directional inequalities that trap

S_{N}

; without monotonicity, a rectangle could sit partly above and partly below the curve and the squeeze collapses.

WHY2. Why does the test tell us convergence but refuse to tell us the sum?

Because it delivers a pair of bounds

\int_{1}^{N + 1} f \leq S_{N} \leq f (1) + \int_{1}^{N} f

, not an equation. The bounds pin down whether

S_{N}

is finite, but leave a genuine gap between them — the sum lives somewhere inside, and the test never says exactly where.

WHY3. Why does the substitution

u = ln x

appear when handling

\sum \frac{1}{n l n n}

Because

d u = d x / x

turns

\frac{d x}{x l n x}

into

\frac{d u}{u}

— the harmonic integral in disguise. The substitution is chosen precisely to reveal that the

ln

factor grows too slowly to fix the divergence.

WHY4. Why is

p = 1

the exact boundary and not

p = 1.5

p = 0.5

Evaluate the improper integral as a limit

\int_{1}^{\infty} x^{- p} d x = lim_{b \to \infty} \int_{1}^{b} x^{- p} d x

, where

b

is just the moving upper cutoff we push to infinity. For

p \neq = 1

this equals

lim_{b \to \infty} \frac{b ^{1 - p} - 1}{1 - p}

, whose fate hinges on the sign of

1 - p

: for

p > 1

the exponent

1 - p < 0

b^{1 - p} \to 0

(finite area), for

p < 1

it's

> 0

b^{1 - p} \to \infty

. At exactly

p = 1

the antiderivative switches to

ln b

, and

lim_{b \to \infty} ln b = \infty

. The behaviour changes character exactly at

1

WHY5. Why can we start the integral test at

n = N

instead of

n = 1

Convergence is a tail property — chopping off finitely many terms changes the sum by a finite amount but never its finite/infinite fate. So if

f

is only eventually positive-decreasing (from some

N

on), test the tail; the head is harmless.

WHY6. Why does the Comparison test use the p-series as its favourite yardstick?

Because p-series have a known, clean convergence rule (

p > 1

), so comparing an unknown series to

1/ n^{p}

instantly settles it. The p-series is the ruler because the integral test already measured it exactly.

Edge cases

EC1. What happens to

\sum 1/ n^{p}

p = 1

exactly?

It is the harmonic series, which diverges — the boundary belongs to the diverging side. "Push past 1" means strictly greater than 1.

EC2. What about

p < 0

, e.g.

\sum n^{2}

Diverges immediately: with

p < 0

the terms

n^{∣ p ∣}

grow without bound, so

a_{n} \neq \to 0

and the n-th term test for divergence kills it before the integral test is even needed.

EC3. What about

p = 0

, i.e.

\sum 1/ n^{0} = \sum 1

Diverges: every term is

1

, the partial sums are

S_{N} = N \to \infty

. Terms don't approach

0

, so n-th term test rules it out instantly.

EC4. If

f

is decreasing but only eventually (say increases then decreases), can we still test?

Yes — apply the test on the tail

[N, \infty)

where

f

is decreasing. The finitely many earlier terms add a finite constant and cannot change convergence.

EC5. Does

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

converging require

f (x) \to 0

No — but for a positive decreasing

f

, if the integral converges then

f (x)

must

\to 0

; otherwise the tail area would be infinite. The direction "integral converges

\Rightarrow f \to 0

" holds under our hypotheses.

EC6. What is the fate of

\sum \frac{1}{n ( l n n ) ^{2}}

compared to

\sum \frac{1}{n l n n}

The extra power flips the result: with

u = ln x

\int_{2}^{\infty} \frac{d x}{x ( l n x ) ^{2}} = \int_{l n 2}^{\infty} \frac{d u}{u ^{2}}

converges (it's a p-integral with

p = 2

), so this series converges while

\sum \frac{1}{n l n n}

diverges. The

ln

acts like a second-order p-series.

Connections

Comparison test — where these p-series yardsticks get used.
Limit comparison test — compares an unknown series against $1/ n^{p}$ .
Improper integrals — the machinery evaluated inside every "why" above.
Harmonic series — the star boundary case $p = 1$ .
Riemann zeta function — $ζ (p) = \sum 1/ n^{p}$ , defined only where the p-series converges.
n-th term test for divergence — the necessary-not-sufficient trap behind TF4, EC2, EC3.