Divergence test (necessary but not sufficient): Question bank

Before we start, one reminder of the vocabulary, so no symbol appears un-earned:

True or false — justify

Terms → 0 guarantees the series converges.

False. That is the converse of the test and it fails:

\sum \frac{1}{n}

has terms

\to 0

yet diverges (see Harmonic Series). Terms → 0 is necessary, not sufficient.

lim a_{n} \neq = 0

then

\sum a_{n}

diverges.

True. This is exactly the test; a non-vanishing term means the running total keeps getting kicked by roughly the same-sized amount forever, so it cannot settle.

The Divergence Test can prove a series converges.

False. It is a one-way street: its only outputs are "diverges" or "inconclusive". It can never emit "converges".

lim a_{n} = 0

the series definitely converges.

False. Zero limit only passes the necessary condition; the series may still diverge (harmonic) or converge (the $p$-series

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

). You must reach for a stronger test.

"Necessary but not sufficient" means every convergent series has terms → 0, but not every terms-→-0 series converges.

True. Convergence forces terms → 0 (necessary), while terms → 0 does not force convergence (not sufficient).

lim a_{n}

does not exist, the series diverges.

True. "DNE" is not equal to

0

, so it triggers the test just like a nonzero limit. Example:

\sum (- 1)^{n}

diverges because

(- 1)^{n}

never settles.

The Divergence Test is the contrapositive of "converges

\Rightarrow

terms

\to 0

True. From

P \Rightarrow Q

(converges

\Rightarrow

limit is 0) we form not-

Q \Rightarrow

not-

P

(limit not 0

\Rightarrow

diverges), which is logically identical.

If two series have the same limit-zero terms, they must have the same fate.

False.

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

converges and

\sum \frac{1}{n}

diverges, yet both have terms

\to 0

. Same limit, opposite fates — proof the test cannot distinguish them.

A series can converge even though infinitely many of its terms are large.

False. If infinitely many terms stay bounded away from

0

, the limit is not

0

(or DNE), so by the test it must diverge; convergence forbids that.

Spot the error

lim \frac{1}{n} = 0

, so by the Divergence Test

\sum \frac{1}{n}

converges."

The error is expecting a "converges" verdict. When the limit is

0

the test is silent; it never outputs convergence. In fact

\sum \frac{1}{n}

diverges — you would need the Integral Test or the doubling-bracket argument to see that.

lim (- 1)^{n} = 0

because the values average out, so the series converges."

Two errors. The limit is not

0

— it does not exist, because the terms alternate

- 1, + 1

forever. And since the limit fails to be

0

, the test concludes the series diverges, not converges.

"The terms of

\sum \frac{3 n ^{2} + 1}{5 n ^{2} + n}

go to

0

, so the test is inconclusive."

The limit is

\frac{3}{5}

, not

0

(divide top and bottom by

n^{2}

). Since

\frac{3}{5} \neq = 0

, the test is decisive: the series diverges. The claim mislabels a nonzero limit as zero.

"I showed

lim a_{n} = 0

, therefore I'm done — the series is classified."

A zero limit finishes nothing. The test returns "inconclusive", so the classification is still open. You must apply another tool (comparison,

p

-series, integral, ratio, alternating).

S_{N} \to 0

, so the series converges to a sum of

0

Confusing the term with the running total.

S_{N}

is the partial sum;

a_{n}

is the individual term. Convergence means

S_{N}

approaches a finite

S

(often nonzero), not that

S_{N}

approaches

0

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

has tiny terms that vanish, so it converges."

Vanishing terms never prove convergence.

\frac{1}{l n n} \to 0

but shrinks slower than

\frac{1}{n}

, so by comparison with the Harmonic Series it diverges. Slow enough shrinking still lets infinitely many pieces pile to infinity.

"Since

a_{N} = S_{N} - S_{N - 1}

and both go to

S

, the difference is

S

, so terms

\to S

Arithmetic slip:

S - S = 0

, not

S

. Both partial sums approach the same limit

S

, so their difference approaches

0

— which is precisely why convergent series have terms → 0.

Why questions

Why is

lim a_{n} = 0

only necessary and not sufficient for convergence?

Because the terms can shrink too slowly: infinitely many tiny pieces can still add to infinity (harmonic series). Shrinking is required to have any hope, but it does not control the rate, which is what actually decides the sum.

Why does the identity

a_{N} = S_{N} - S_{N - 1}

sit at the heart of the derivation?

It bridges the term and the running total: "the new term = total now minus total one step ago". Taking limits then forces

a_{N} \to S - S = 0

whenever the series converges. The same telescoping identity powers Telescoping Series.

Why does "the limit does not exist" count the same as "the limit is nonzero"?

The test needs the limit to equal $0$ to stay silent. Any failure of that — a nonzero value or no value at all — means the terms don't shrink to zero, so the necessary condition is violated and the series diverges.

Why is the Divergence Test called the cheapest first check?

It costs one limit of

a_{n}

(via Limits of Sequences) and can instantly kill many series before any harder machinery. If the limit is clearly nonzero, you are done in a single line.

Why can't we conclude convergence from grouping the harmonic series into brackets

\geq \frac{1}{2}

That grouping shows the opposite — each doubling block sums to at least

\frac{1}{2}

, so adding infinitely many blocks pushes the total to infinity. It is a divergence argument, demonstrating that terms → 0 is not enough.

Why does the test never fire on a

p

-series with

p = 1

versus

p = 2

Both have terms

\frac{1}{n ^{p}} \to 0

, so the limit condition is passed in each case and the test stays silent. Distinguishing

p = 1

(diverges) from

p = 2

(converges) requires the p-Series Test or Integral Test.

Edge cases

What does the test say about

\sum_{n = 1}^{\infty} (1 + \frac{1}{n})^{n}

The terms tend to

e \approx 2.718 \neq = 0

, so the necessary condition fails and the series diverges. A limit that looks like it might approach

1

actually approaches

e

What about a constant nonzero series

\sum 5

lim 5 = 5 \neq = 0

, so it diverges immediately — you are adding

5

forever and the running total climbs without bound.

What about

\sum 0

, the all-zeros series?

lim a_{n} = 0

, so the test is inconclusive by rule; but here the partial sums are all

0

, so the series trivially converges to

0

. A reminder that "inconclusive" does not mean "diverges".

What if only the tail of the terms matters — does an early spike change the verdict?

No. Convergence and

lim a_{n}

both depend only on the long-run behaviour, so changing finitely many early terms cannot alter the test's conclusion.

What does the test say about

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

The terms tend to

1 \neq = 0

, so it diverges. Even though each term is less than

1

, they never shrink toward

0

What about

\sum sin n

(angle in radians)?

lim sin n

does not exist — the values wander densely over

[- 1, 1]

and never settle. Since the limit is not

0

, the test declares divergence.

If the terms oscillate but with shrinking amplitude, say

a_{n} = \frac{( - 1 ) ^{n}}{n}

, what does the test say?

lim a_{n} = 0

, so the test is silent. The alternating structure is decided elsewhere (it actually converges) — the Divergence Test gives no information here.

Connections

Parent: Divergence Test — the full derivation these traps target.
Partial Sums and Series Convergence — the $S_{N}$ vs $a_{n}$ distinction behind the "spot the error" set.
Harmonic Series — the canonical "terms → 0 yet diverges" counterexample.
p-Series Test, Integral Test, Comparison Test — the tools that resolve the inconclusive ( $lim a_{n} = 0$ ) cases.
Telescoping Series — uses the $a_{N} = S_{N} - S_{N - 1}$ identity directly.
Limits of Sequences — the engine for computing $lim a_{n}$ .