Limit comparison test: Question bank — Maths deep dive

Deep dive child of Limit Comparison Test (4.3.9) · also read the Hinglish version.

These are conceptual traps — no heavy algebra. Each one targets a place where students think they understand the Limit Comparison Test (LCT) but a boundary case or a hidden assumption bites them. Cover the answer, reason it out in a sentence, then reveal.

True or false — justify

True or false: If

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

, then

\sum a_{n}

converges.

False.

a_{n} \to 0

is only necessary, not sufficient — the harmonic series

\sum 1/ n

has terms going to

0

yet diverges because they shrink too slowly.

True or false: In the LCT the terms

a_{n}

and

b_{n}

must be positive.

True. The proof multiplies a ratio inequality by

b_{n}

without flipping it (needs

b_{n} > 0

) and leans on the Direct Comparison Test, which itself requires positive terms.

True or false: If

0 < L < \infty

then

\sum a_{n}

always converges.

False.

0 < L < \infty

only means the two series share a fate — they either both converge or both diverge. You still must know whether

\sum b_{n}

converges to conclude anything.

True or false: If

L = 0

, then

\sum a_{n}

converges.

False.

L = 0

only helps when

\sum b_{n}

converges (then

a_{n}

is eventually smaller and gets dragged down). If

\sum b_{n}

diverges,

L = 0

tells you nothing.

True or false: If

L = \infty

, then

\sum a_{n}

diverges.

False.

L = \infty

only helps when

\sum b_{n}

diverges (then

a_{n}

is eventually bigger and pushed up). If

\sum b_{n}

converges,

L = \infty

is inconclusive.

True or false: If

L = lim a_{n} / b_{n}

does not exist, the LCT simply fails to apply in its main form.

True. The main test needs a genuine limit

L

; if the ratio oscillates forever there is no single

L

, so you must use a different tool such as the Direct Comparison Test or Ratio Test.

True or false: Swapping the roles of

a_{n}

and

b_{n}

(computing

b_{n} / a_{n}

instead) can change which conclusions you may draw.

True. In the main case (

0 < L < \infty

) nothing changes, since

1/ L

is again finite and positive. But in the one-sided cases it flips:

lim a_{n} / b_{n} = 0

becomes

lim b_{n} / a_{n} = \infty

, so a card that lets you inherit convergence turns into one that can only inherit divergence — the usable direction swaps.

True or false: The LCT can be used on

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

False directly, because the terms are not all positive. You'd instead test

\sum ∣ a_{n} ∣ = \sum 1/ n

(which diverges) or use the Alternating Series Test on the original.

Spot the error

A student writes: "

\sum \frac{n + 1}{n ^{3} + n}

behaves like

\frac{1}{n ^{3}}

, so

b_{n} = 1/ n^{3}

." Where's the slip?

They dropped the numerator's growth. Keep the strongest power top and bottom separately: top

\sim n

, bottom

\sim n^{3}

, so

b_{n} \sim n / n^{3} = 1/ n^{2}

A student picks

b_{n} = 1/ n

for

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

and computes

L

. What actually happens, and why is

1/ n

a poor choice?

The ratio is

\frac{( ln n ) / n ^{2}}{1/ n} = \frac{ln n}{n} \to 0

, so

L = 0

(not

\infty

— the extra

n

in the denominator beats the slow

ln n

). But

\sum 1/ n

diverges, and

L = 0

with a divergent

b_{n}

is inconclusive. Switch to

b_{n} = 1/ n^{1.5}

, still giving

L = 0

but now against a convergent series, so you can conclude convergence.

A student says: "

a_{n} < b_{n}

eventually and

\sum b_{n}

diverges, so by LCT

\sum a_{n}

diverges." Fix it.

Being smaller than a divergent series proves nothing (the smaller one might converge). Divergence transfers only from a smaller divergent series to a larger one, i.e. you need

a_{n} > b_{n}

with

\sum b_{n}

divergent.

A student compares

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

with

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

and gets a limit, then says "converges." Spot the error.

The ratio is

\frac{1/ n ^{2} + 1}{1/ n ^{2}} = \frac{n ^{2}}{n ^{2} + 1} \approx n \to \infty

, so

L = \infty

— not

0

. And

L = \infty

against a convergent

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

is inconclusive, so no verdict follows. The right skeleton is

b_{n} = 1/ n

⇒

L = 1

⇒ diverges.

A student writes

L = lim \frac{a _{n}}{b _{n}} = \frac{l i m a _{n}}{l i m b _{n}} = \frac{0}{0}

and stops. What's the mistake?

You cannot split the limit of a ratio into a ratio of limits when both go to

0

— that's the indeterminate form

0/0

. You must simplify the ratio itself first (factor, divide by the top power) before taking the limit.

A student claims

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

needs the Ratio Test because of the

2^{n}

. Is anything wrong?

Not wrong, but LCT is cleaner: for large

n

the

2^{n}

dominates, so

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

(a geometric series with

r = \frac{1}{2}

) gives

L = 1

and convergence immediately.

Why questions

Why do we choose

ε = L /2

in the proof rather than any small

ε

The limit definition says: for any

ε > 0

there is an

N

with

∣ a_{n} / b_{n} - L ∣ < ε

for

n > N

, i.e.

L - ε < a_{n} / b_{n} < L + ε

. Picking

ε = L /2

makes the lower end

L - ε = L /2

, which is strictly positive since

L > 0

; any bigger

ε

could drop the lower bound to

0

or below and kill the useful inequality

\frac{L}{2} b_{n} < a_{n}

Why does the LCT only care about "large

n

" and ignore the first few terms?

Convergence is a tail property — adding or changing finitely many terms shifts the sum by a finite amount but cannot turn a convergent series divergent or vice versa. The limit

L

reads only the tail behaviour.

Why does the mnemonic "keep the strongest, ditch the rest" work for building

b_{n}

For huge

n

the highest-power term dwarfs all weaker ones, so the ratio of leading terms captures the true rate of growth/decay — the only thing that decides convergence.

Why is the LCT often preferable to the Direct Comparison Test?

Direct comparison demands an exact inequality

a_{n} \leq b_{n}

(or

\geq

) in the right direction, which can be painful or point the wrong way; LCT only needs the two terms to shrink at the same rate, sidestepping the inequality entirely.

Why is

\sum 1/ n^{2} + 1

a textbook case for the LCT rather than direct comparison?

Because

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

, which is the wrong direction to prove divergence by direct comparison (smaller than a divergent series proves nothing). LCT gives

L = 1

against

1/ n

and settles it.

Why does

L = \infty

with a divergent

b_{n}

actually force

\sum a_{n}

to diverge?

L = \infty

means for any big

M

there is an

N

with

a_{n} / b_{n} > M

for

n > N

, i.e.

a_{n} > M b_{n}

. Since

\sum b_{n}

diverges, so does

\sum M b_{n}

, and

a_{n}

sits above it term-by-term — by the Direct Comparison Test

\sum a_{n}

diverges too.

Why does knowing the fate of [[p-Series|

p

-series]] and Geometric Series matter so much for the LCT?

The LCT never decides on its own — it transfers a known verdict from

b_{n}

a_{n}

. So you need a stockpile of series whose fate you already know, and

p

-series and geometric series are the two workhorses.

Edge cases

a_{n} = b_{n}

for all

n

, what is

L

and what does the test say?

L = 1

, safely inside

0 < L < \infty

, so it says "same fate" — trivially true since they are the same series. Harmless but content-free.

Suppose

a_{n} / b_{n}

oscillates between

1

and

2

forever (no limit). Can you still salvage LCT?

Not the main limit form, but the ratio stays trapped in

[1, 2]

, so you can bound

b_{n} \leq a_{n} \leq 2 b_{n}

and fall back on the Direct Comparison Test or the Squeeze Theorem on the ratio.

What happens if

\sum b_{n}

is itself a borderline series like the harmonic

\sum 1/ n

Nothing special breaks — LCT still transfers the (divergent) verdict of

\sum 1/ n

\sum a_{n}

whenever

0 < L < \infty

. The borderline nature of

b_{n}

is exactly why you must pick

b_{n}

's exponent carefully.

If both

\sum a_{n}

and

\sum b_{n}

converge, must

0 < L < \infty

No. Two convergent series can decay at wildly different rates, giving

L = 0

(e.g.

a_{n} = 1/ n^{3}

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

) or

L = \infty

. A finite positive

L

is sufficient for shared fate but not necessary.

What is the verdict if

L = 0

and

\sum b_{n}

diverges?

Inconclusive.

a_{n}

is eventually smaller than

b_{n}

, but being smaller than a divergent series says nothing —

\sum a_{n}

could go either way. Choose a different

b_{n}

What is the verdict if

L = \infty

and

\sum b_{n}

converges?

Inconclusive.

a_{n}

is eventually larger than

b_{n}

, but being larger than a convergent series says nothing about

\sum a_{n}

. Pick a

b_{n}

that yields

0 < L < \infty

Can the LCT ever prove a series diverges using a convergent

b_{n}

Only in the

L = \infty

mismatch, where it can't — so effectively no. To prove divergence you need either

0 < L < \infty

with

\sum b_{n}

divergent, or

L = \infty

with

\sum b_{n}

divergent.

Recall Quick self-test before you leave

$L = 0$ helps only when $\sum b_{n}$ ::: converges. $L = \infty$ helps only when $\sum b_{n}$ ::: diverges. $0 < L < \infty$ means the two series ::: share a fate (both converge or both diverge). The single required condition on the terms is ::: both $a_{n} > 0$ and $b_{n} > 0$ for all large $n$ .