Direct comparison test: Question bank — Maths deep dive

Two ideas power almost every answer here, so let us pin them down in plain words first.

Recall The two arrows (the whole game)

Convergence flows DOWN: to trap a series under a finite roof, you need a bigger series that is known to converge. If your series sits below something finite, it too is finite. Divergence flows UP: to push a series above an infinite floor, you need a smaller series that is known to diverge. If your series sits above something infinite, it too is infinite. Anything else — a smaller convergent partner, or a bigger divergent partner — tells you nothing.

The picture below is the mental image behind every answer on this page — refer back to it whenever an arrow feels ambiguous.

Recall Why the roof/floor actually forces the conclusion (the engine)

The two arrows are not magic — they run on the Monotone convergence theorem. Look at the figure above: build the partial sums $S_{n} = a_{1} + \dots + a_{n}$ . Because every term $a_{n} \geq 0$ , each new term only adds height, so $S_{n}$ can never step down — it is an increasing staircase (blue). If a convergent $\sum b_{n}$ sits above it, that staircase is also capped by the finite total $B$ (yellow dashed roof). An increasing staircase that can never rise past a fixed ceiling has nowhere to go but settle — that is Monotone Convergence, and that is why "below a finite roof" means "converges."

True or false — justify

The word "eventually" means "for all $n \geq N$ for some fixed starting index $N$ " — finitely many early terms never affect convergence.

0 \leq a_{n} \leq b_{n}

for all

n

and

\sum b_{n}

converges, then

\sum a_{n}

converges.

True — the convergence half of DCT. The partial sums

S_{n}

a_{n}

are increasing (each term

\geq 0

adds height, never subtracts) and capped by the finite total of

b_{n}

; an increasing-and-capped staircase must settle by the Monotone convergence theorem, so

\sum a_{n}

converges.

0 \leq a_{n} \leq b_{n}

and

\sum a_{n}

converges, then

\sum b_{n}

converges.

False — knowing the small one is finite says nothing about the big one. The bigger series could still blow up;

a_{n} = 1/ n^{2}

converges but

b_{n} = 1/ n

diverges even though

a_{n} \leq b_{n}

0 \leq a_{n} \leq b_{n}

and

\sum a_{n}

diverges, then

\sum b_{n}

diverges.

True — the divergence half. The partial sums of

b_{n}

dominate those of

a_{n}

term-by-term, and

a_{n}

's already run off to infinity, so the bigger staircase is dragged to infinity too (no finite cap can exist).

\sum a_{n}

converges and

\sum b_{n}

converges, DCT was the tool that proved it.

False — DCT is one possible tool, not the only one. Convergence can also come from the ratio, root, integral, or alternating-series tests; DCT specifically needs a term-by-term inequality against a known benchmark.

DCT can be applied to

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

directly.

False — the terms change sign, so the partial-sum staircase can step down, wrecking the monotonicity that the Monotone convergence theorem needs. Compare absolute values instead (see Absolute convergence) or use the Alternating Series Test.

a_{n} \leq b_{n}

for all

n

" is stronger than DCT actually requires.

True — DCT only needs the inequality eventually (for

n \geq N

). Finitely many terms are a finite sum and cannot change whether the tail converges.

lim_{n \to \infty} a_{n} = 0

, then

\sum a_{n}

converges by comparison.

False — terms going to zero is necessary but not sufficient. The Harmonic series

\sum 1/ n

has terms tending to

0

yet diverges; DCT still needs a genuine convergent upper bound.

Showing

a_{n} \leq \frac{1}{2 ^{n}}

(with

a_{n} \geq 0

) proves

\sum a_{n}

converges.

True —

\sum 1/ 2^{n}

is a convergent Geometric series (ratio

1/2 < 1

), a bigger series that caps

a_{n}

's increasing partial sums under a finite roof, so Monotone Convergence forces settling. Correct direction.

Showing

a_{n} \geq \frac{1}{2 ^{n}}

proves

\sum a_{n}

converges.

False — that is a smaller convergent partner, the useless direction. Sitting above something finite gives no ceiling on the staircase;

a_{n}

could still diverge.

Spot the error

Each argument reaches a conclusion using a real inequality. Find where the logic, not the algebra, goes wrong. The figure below shows the two useless directions in one glance.

"To show

\sum \frac{1}{n ^{2} + 1}

converges I note

\frac{1}{n ^{2} + 1} \geq \frac{1}{2 n ^{2}}

, and

\sum \frac{1}{2 n ^{2}}

converges, so done."

Wrong direction — the partner is smaller and convergent, which proves nothing. Use

\frac{1}{n ^{2} + 1} \leq \frac{1}{n ^{2}}

(a bigger convergent series) instead; that traps it from above.

"To show

\sum \frac{1}{n}

diverges I note

\frac{1}{n} \leq \frac{1}{1} = 1

, and comparing to the divergent constant series

\sum 1

finishes it."

Wrong direction —

\frac{1}{n}

is below

\sum 1

, and being smaller than a divergent series proves nothing. Instead use

\frac{1}{n} \geq \frac{1}{n}

(

p = 1/2 \leq 1

vs harmonic), a smaller divergent partner pushing up.

"Since

\frac{2 + s i n n}{n ^{2}}

has a numerator that is sometimes negative-ish, DCT cannot apply."

The premise is false:

- 1 \leq sin n \leq 1

gives

1 \leq 2 + sin n \leq 3

, always positive. Terms are non-negative, so DCT applies fine with the bound

\leq \frac{3}{n ^{2}}

\frac{1}{n ^{2} - 10} \leq \frac{1}{n ^{2}}

for all

n

, so

\sum \frac{1}{n ^{2} - 10}

converges."

The stated inequality is backwards: for every

n

with

n^{2} > 10

(i.e.

n \geq 4

) the denominator

n^{2} - 10

is smaller than

n^{2}

, so

\frac{1}{n ^{2} - 10} > \frac{1}{n ^{2}}

— a bigger series, the useless direction for convergence. (For

n \leq 3

the term is even negative.) To fix it, bound with a slightly looser convergent roof such as

\frac{1}{n ^{2} - 10} \leq \frac{2}{n ^{2}}

for

n

large enough (here

n^{2} - 10 \geq n^{2} /2 ⟺ n^{2} \geq 20 ⟺ n \geq 5

), then

\sum \frac{2}{n ^{2}}

converges.

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

converges because

\frac{l n n}{n} \leq \frac{n}{n} = 1

Bounding term-by-term by a constant

1

is useless —

\sum 1

diverges, and being below a divergent series says nothing. The right move:

ln n \geq 1 ⟺ n \geq e \approx 2.718

, so eventually (for

n \geq 3

)

\frac{l n n}{n} \geq \frac{1}{n}

, a smaller divergent partner — hence it diverges.

\sum a_{n}

with

a_{n} = \frac{c o s ^{2} n}{n ^{2}}

— since

cos^{2} n

oscillates, no clean comparison exists, so the test is inconclusive."

False despair —

0 \leq cos^{2} n \leq 1

gives

0 \leq a_{n} \leq \frac{1}{n ^{2}}

, a clean bigger convergent bound. Oscillation in a bounded numerator is never an obstacle; bound it by its maximum.

Why questions

Why does DCT insist the terms be non-negative?

Only then are the partial sums monotone increasing, which is the exact hypothesis the Monotone convergence theorem needs to conclude a bounded increasing sequence converges. Sign changes break monotonicity and the whole proof.

Why does knowing the smaller series converges give no information about the bigger one?

A finite lower structure cannot cap something above it; the bigger series has extra "room" to grow. Only an upper bound (a bigger convergent series) can force finiteness.

Why is the

p

-series with

p = 2

the go-to convergent benchmark?

By the p-series test,

\sum 1/ n^{p}

converges exactly when

p > 1

, and

1/ n^{2}

is simple, positive, and dominates the tails of many rational-term series — an easy roof to reach for.

Why is the Harmonic series the canonical divergent benchmark?

It is the borderline

p = 1

case that just barely diverges, so any series whose terms are eventually

\geq 1/ n

is pushed above an infinite floor — the cleanest divergence lever.

Why does DCT never tell us the value of the sum?

It only compares behaviour (finite vs infinite) term-by-term; the actual total depends on all the terms exactly, which an inequality deliberately throws away. It answers "does it converge?" not "to what?".

Why can we ignore finitely many misbehaving early terms?

A finite sum is always finite, so it cannot change whether the infinite tail converges. Convergence is a property of the tail, hence "eventually

n \geq N

" is enough.

Why prefer the Limit comparison test when the inequality is awkward?

DCT demands you prove a clean term-by-term inequality; if

a_{n}

and

b_{n}

merely behave alike but you cannot pin the direction, the limit-comparison ratio sidesteps the inequality entirely while giving the same conclusion.

Why is the Integral test sometimes used alongside DCT?

It is an alternate way to classify the benchmark series (like

\sum 1/ n^{p}

) as convergent or divergent, giving you the trusted partner that DCT then compares against.

Edge cases

What does DCT conclude if

a_{n} = 0

for infinitely many

n

Nothing breaks — zeros are non-negative and satisfy

0 \leq a_{n} \leq b_{n}

trivially. Zero terms only help; the comparison and monotonicity arguments are unaffected.

a_{n} \leq b_{n}

but

\sum b_{n}

diverges, what can DCT say about

\sum a_{n}

Nothing — the bigger series diverging leaves the smaller one free to converge or diverge. You have the wrong-direction pairing for divergence and no upper bound for convergence.

Series has terms

\geq 0

but the natural bound only holds for even

n

, failing for odd

n

— is DCT usable?

Not directly, since DCT needs the inequality for all

n \geq N

, not a subsequence. You must find a single bound valid for the whole tail (or split and handle each piece).

Both

a_{n} \geq 0

and

a_{n} \leq b_{n}

hold, and

\sum a_{n}

converges — does that force

b_{n} \to 0

No — the bigger series

b_{n}

is entirely unconstrained by the small one converging.

b_{n}

need not even tend to

0

; DCT gives no downward information.

A series has

a_{n} \to 0

and is squeezed as

\frac{1}{n ^{2}} \leq a_{n} \leq \frac{1}{n}

— convergent or divergent?

Inconclusive from these bounds alone: the lower (

1/ n^{2}

) converges and the upper (

1/ n

) diverges, so DCT gives no verdict. Neither trap fires; try Limit comparison test with the true dominant term of

a_{n}