Divergence test (necessary but not sufficient): Worked examples

Before anything, recall the only two verdicts this test can ever return:

Recall The rule in one breath

Compute $L = n \to \infty lim a_{n}$ (the term you're adding, as $n$ grows huge). If $L \neq = 0$ or the limit does not exist (DNE), the series diverges. If $L = 0$ , the test is inconclusive — it says nothing, and you must reach for another tool.

Here $a_{n}$ just means "the $n$ -th number in the list you are summing", and $lim_{n \to \infty} a_{n}$ asks "as $n$ marches off to infinity, does that number settle onto a single value, and if so, which?" See Limits of Sequences — that is the engine that powers everything below.

The scenario matrix

Every series you can feed the Divergence Test lands in exactly one of these cells. The right column is the verdict the Divergence Test alone gives.

Cell	What the term $a_{n}$ does as $n \to \infty$	Test verdict	Example that hits it
A	Settles on a nonzero number (e.g. rational function, equal degrees)	DIVERGES	Ex 1
B	Grows without bound ( $a_{n} \to \pm \infty$ )	DIVERGES	Ex 2
C	Oscillates, never settling (limit DNE)	DIVERGES	Ex 3
D	Approaches a nonzero constant via a famous limit ( $\to e$ )	DIVERGES	Ex 4
E	Goes to zero slowly → still diverges (test blind)	INCONCLUSIVE	Ex 5
F	Goes to zero fast → actually converges (test blind)	INCONCLUSIVE	Ex 6
G	Sign flips but shrinks to zero (alternating)	INCONCLUSIVE	Ex 7
H	Word problem — pebbles/medicine doses, must extract $a_{n}$	depends	Ex 8
I	Exam twist — looks like it decays but a hidden constant survives	DIVERGES	Ex 9

Figure — Divergence test (necessary but not sufficient)

Look at the red curve on the left: its terms level off above zero — the test fires. On the right the terms all crawl down to zero, and the test cannot tell the two apart even though one series adds up to a finite number and the other blows up.

Cell A — term settles on a nonzero number

Worked example Example 1 (Cell A) — equal-degree rational terms

$\sum_{n = 1}^{\infty} \frac{4 n ^{3} - 2 n}{7 n ^{3} + n ^{2} + 5} .$ Forecast: cover the answer and guess. Top and bottom both have $n^{3}$ as their fastest-growing piece. What number does a ratio of two things that grow at the same rate settle on? Not zero — a plain ratio of the leading coefficients. So predict: diverges.

Step 1. Divide numerator and denominator by $n^{3}$ , the highest power present. $a_{n} = \frac{4 n ^{3} - 2 n}{7 n ^{3} + n ^{2} + 5} = \frac{4 - \frac{2}{n ^{2}}}{7 + \frac{1}{n} + \frac{5}{n ^{3}}} .$ Why this step? Dividing by the fastest-growing term turns every slower term into something with $n$ in its denominator, which we can then send to zero. It exposes who dominates.

Step 2. Send $n \to \infty$ . Every fraction with $n$ downstairs dies: $L = \frac{4 - 0}{7 + 0 + 0} = \frac{4}{7} .$ Why this step? $\frac{2}{n ^{2}} \to 0$ , $\frac{1}{n} \to 0$ , $\frac{5}{n ^{3}} \to 0$ — a constant over a growing denominator vanishes.

Step 3. Since $L = \frac{4}{7} \neq = 0$ , the Divergence Test fires: the series diverges.

Verify: sanity-check with a large $n = 1000$ : $\frac{4 ( 1 0 ^{9} ) - 2000}{7 ( 1 0 ^{9} ) + 1 0 ^{6} + 5} \approx 0.5714 \approx \frac{4}{7}$ . Terms hover near $0.57$ forever, so adding infinitely many of them can't total a finite number. ✔

Cell B — term grows without bound

Worked example Example 2 (Cell B) — numerator wins

$\sum_{n = 1}^{\infty} \frac{n ^{2} + 1}{3 n + 4} .$ Forecast: top has degree 2, bottom degree 1. The top out-grows the bottom, so the term itself blows up. Predict: diverges (even harder than Cell A — the terms don't just stay big, they explode).

Step 1. Divide by $n$ (the highest power in the denominator, so we can read the growth): $a_{n} = \frac{n ^{2} + 1}{3 n + 4} = \frac{n + \frac{1}{n}}{3 + \frac{4}{n}} .$ Why this step? This isolates the surviving $n$ upstairs, showing the term grows like $\frac{n}{3}$ .

Step 2. As $n \to \infty$ : numerator $\to \infty$ , denominator $\to 3$ , so $L = lim_{n \to \infty} \frac{n + \frac{1}{n}}{3 + \frac{4}{n}} = + \infty.$ Why this step? A quantity heading to infinity divided by a finite constant still heads to infinity.

Step 3. $L = + \infty \neq = 0$ , so the series diverges.

Verify: at $n = 100$ , $a_{100} = \frac{10001}{304} \approx 32.9$ ; at $n = 1000$ , $a_{1000} \approx 333.1$ — the terms keep climbing, confirming $L = \infty$ . ✔

The red points march upward off the top of the plot — a term that grows can never let a running total settle.

Cell C — term oscillates (limit DNE)

Worked example Example 3 (Cell C) — bouncing sign

$\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n} n}{n + 1} .$ Forecast: ignore the sign for a second — $\frac{n}{n + 1} \to 1$ . But the $(- 1)^{n}$ makes the term flip between about $+ 1$ and $- 1$ . Does that settle on one value? No. Predict: diverges because the limit does not exist.

Step 1. Split the magnitude from the sign. The size is $\frac{n}{n + 1} = \frac{1}{1 + \frac{1}{n}} \to 1$ . Why this step? Knowing the size approaches $1$ (not $0$ ) tells us the flips are between numbers near $- 1$ and $+ 1$ , not between shrinking numbers.

Step 2. Track even and odd $n$ separately:

even $n$ : $a_{n} \to + 1$ ,
odd $n$ : $a_{n} \to - 1$ .

Why this step? If two subsequences aim at two different values, the whole sequence has no single limit — the limit does not exist.

Step 3. DNE counts as " $\neq = 0$ " for this test, so the series diverges.

Verify: $a_{100} = \frac{100}{101} \approx + 0.990$ , $a_{101} = - \frac{101}{102} \approx - 0.990$ . The terms straddle $\pm 1$ forever — no settling, no finite total. ✔

Cell D — famous nonzero limit

Both red term-curves flatten onto zero (top panel), indistinguishable to the test. The bottom panel shows the running totals: one keeps climbing (diverges), the other flattens onto $\frac{π ^{2}}{6}$ (converges). The test only sees the top panel — that's why it can't tell them apart.

Cell G — alternating, shrinks to zero

Worked example Example 7 (Cell G) — the sign flips but the size dies

$\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n + 1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots .$ Forecast: the magnitude $\frac{1}{n} \to 0$ , so the term (with sign) also goes to $0$ . Predict: the Divergence Test is inconclusive.

Step 1. $∣ a_{n} ∣ = \frac{1}{n} \to 0$ , and squeezing gives $a_{n} = \frac{( - 1 ) ^{n + 1}}{n} \to 0$ . Why this step? A quantity trapped between $- \frac{1}{n}$ and $+ \frac{1}{n}$ , both of which go to $0$ , is forced to $0$ as well.

Step 2. $L = 0$ , so the Divergence Test says nothing.

Step 3. A separate tool (the alternating-series argument) shows this one actually converges — to $ln 2$ . Compare with Cell E's $\sum \frac{1}{n}$ : same-sized terms, but the alternating signs let cancellation save it.

Verify: $ln 2 \approx 0.6931$ ; the partial sum $S_{1000} \approx 0.6926$ , homing in on $ln 2$ . Zero-limit terms ⇒ test silent ⇒ we needed the other tool. ✔

Cell H — word problem (extract $a_{n}$ first)

Worked example Example 8 (Cell H) — the never-shrinking dose

A patient takes a pill. Each hour their body keeps a fixed fraction so that the drug present from the $n$ -th dose, measured hours later, contributes $a_{n} = 5 \cdot \frac{n}{n + 3}$ mg to a running "total exposure" tally. Does the total exposure $\sum a_{n}$ settle to a finite number? Forecast: each contribution approaches $5 \cdot \frac{n}{n + 3} \to 5$ mg — it does not shrink. Predict: total exposure diverges (grows without bound).

Step 1. Extract the term: $a_{n} = 5 \cdot \frac{n}{n + 3}$ . Divide by $n$ : $a_{n} = 5 \cdot \frac{1}{1 + \frac{3}{n}}$ . Why this step? The word problem hides an ordinary sequence; we must first name $a_{n}$ before any test applies.

Step 2. $L = 5 \cdot \frac{1}{1 + 0} = 5 \neq = 0$ (units: mg).

Step 3. $L = 5 mg \neq = 0$ , so by the Divergence Test the total exposure diverges — physically, contributions never taper off, so cumulative exposure grows without bound.

Verify: $a_{1000} = 5 \cdot \frac{1000}{1003} \approx 4.985$ mg — still near $5$ , confirming a nonzero limit and unbounded total. ✔

Cell I — exam twist (hidden surviving constant)

Worked example Example 9 (Cell I) — looks like it decays, but…

$\sum_{n = 1}^{\infty} (n^{2} + n - n) .$ Forecast: at a glance $n^{2} + n \approx n$ , so the term looks like it might vanish. But does the leftover really go to zero? Predict carefully — this is the trap.

Step 1. Multiply by the conjugate to tame the subtraction of two huge nearly-equal roots: $a_{n} = \frac{( n ^{2} + n - n ) ( n ^{2} + n + n )}{n ^{2} + n + n} = \frac{n ^{2} + n - n ^{2}}{n ^{2} + n + n} = \frac{n}{n ^{2} + n + n} .$ Why this step? Subtracting two large near-equal numbers loses all precision; the conjugate trick converts it into a clean quotient we can take a limit of.

Step 2. Divide top and bottom by $n$ : $a_{n} = \frac{1}{1 + \frac{1}{n} + 1} ⟶ \frac{1}{1 + 1} = \frac{1}{2} .$ Why this step? Pulling out $n$ from the root uses $n^{2} + n = n 1 + \frac{1}{n}$ ; then $\frac{1}{n} \to 0$ .

Step 3. $L = \frac{1}{2} \neq = 0$ — the "decay" was an illusion. The series diverges.

Verify: at $n = 1000$ , $1000000 + 1000 - 1000 \approx 1000.4998 - 1000 = 0.4998 \approx \frac{1}{2}$ . The gap parks at $\frac{1}{2}$ , never zero. ✔

Recap of every cell

Recall Self-test: name the verdict for each

Cell A (nonzero settle) ::: DIVERGES Cell B (grows to infinity) ::: DIVERGES Cell C (oscillates, limit DNE) ::: DIVERGES Cell D (famous $\to e^{x}$ ) ::: DIVERGES Cell E ( $\to 0$ slowly, e.g. $1/ n$ ) ::: INCONCLUSIVE (and actually diverges) Cell F ( $\to 0$ fast, e.g. $1/ n^{2}$ ) ::: INCONCLUSIVE (and actually converges) Cell G (alternating $\to 0$ ) ::: INCONCLUSIVE (and actually converges to $ln 2$ ) Cell I (hidden surviving constant) ::: DIVERGES

Connections

Divergence test — Hinglish parent — the rule these examples exercise.
Limits of Sequences — every "compute $L$ " step lives here.
Partial Sums and Series Convergence — the running-total picture in Cell F's figure.
Harmonic Series — Cell E's star blind-spot example.
p-Series Test — resolves Cells E and F that the test cannot.
Integral Test, Comparison Test — the "next tools" when $L = 0$ .
Telescoping Series — where the $a_{N} = S_{N} - S_{N - 1}$ idea returns.

Worked examples — Divergence test (necessary but not sufficient)