Chapter 12

Medium

Convergence Tests

00 · Symbol Glossary

$\displaystyle\sum \frac{1}{n^p}$p-series

A series with terms $\dfrac{1}{n^p}$ for a fixed exponent $p$ . The series analogue of Chapter 08's p-integral $\int_1^\infty x^{-p}\,dx$ — Section 02 shows the same threshold value of $p$ decides convergence for both.

$0\leq a_n \leq b_n$term-by-term comparison

A relationship between two series' terms, both non-negative, used in the comparison test (Section 01) to transmit convergence or divergence from one series to the other.

$\displaystyle\sum (-1)^{n-1} b_n$alternating series

A series whose terms strictly alternate in sign, with $b_n>0$ for every $n$ . The sign alternation itself can force convergence even when $\sum b_n$ (without the alternating signs) diverges — Section 04's central example.

$\text{abs. conv. / cond. conv.}$absolute and conditional convergence

Two different strengths of convergence for a series with mixed-sign terms, distinguished in Section 05: whether the series of absolute values also converges, or only the original signed series does.

01 · The Comparison Test

The geometric series (Chapter 09) and p-series (Section 02) are two rare series whose convergence can be checked directly. Most other series are checked instead by comparing them, term by term, to one of these known cases.

Definition — Comparison Test

Suppose $0 \leq a_n \leq b_n$ for all $n$ beyond some starting index.

If $\sum b_n$ converges, then $\sum a_n$ converges (a smaller non-negative running total, capped by a finite one, cannot itself grow without bound).
If $\sum a_n$ diverges, then $\sum b_n$ diverges (a larger running total than one already growing without bound must itself grow without bound).

Example — Comparing to a known geometric series

Does $\displaystyle\sum_{n=1}^{\infty}\frac{1}{2^n+n}$ converge? For every $n\geq1$ : $2^n+n > 2^n$ , so $\dfrac{1}{2^n+n} < \dfrac{1}{2^n}$ (a larger denominator gives a smaller fraction, both sides positive).

$\displaystyle\sum\frac{1}{2^n} = \sum\left(\frac12\right)^n$ is geometric with $r=\frac12$ , $\lvert r\rvert<1$ , so it converges (Chapter 09, Section 03). Since $0\leq\frac{1}{2^n+n}<\frac{1}{2^n}$ and the larger series converges, the comparison test gives that $\displaystyle\sum\frac{1}{2^n+n}$ converges too.

02 · p-Series and the Integral Test

Definition — p-Series Test

\sum_{n=1}^{\infty} \frac{1}{n^p} \text{ converges if } p>1, \quad \text{diverges if } p\leq1

Definition — Integral Test

If $f(x)$ is positive, continuous, and decreasing for $x\geq1$ , and $a_n=f(n)$ , then $\displaystyle\sum_{n=1}^{\infty} a_n$ and $\displaystyle\int_1^\infty f(x)\,dx$ (Chapter 08) either both converge or both diverge.

Why the integral test works — geometric intuition

Drawing a rectangle of width $1$ and height $f(n)$ over each interval $[n,n+1]$ produces exactly the Riemann-sum picture of Chapter 06, Section 01. Since $f$ is decreasing, these rectangles either all sit above the curve or all sit below it (depending on which endpoint is used), sandwiching the series' partial sums between the integral and the integral plus one extra term — so one is finite exactly when the other is.

Step-by-step — Prove the p-series test using the integral test, for p ≠ 1

Set $f(x) = x^{-p}$ , matching the p-series' terms $a_n = n^{-p}$ :

f

is positive, continuous, and decreasing for

x\geq1

whenever

p>0

(larger

x

gives a smaller

x^{-p}

Apply the integral test:

\sum n^{-p}

converges if and only if

\int_1^\infty x^{-p}\,dx

converges.

Recall Chapter 08, Section 03's p-integral result directly:

\int_1^\infty x^{-p}\,dx

converges iff

p>1

Conclude by the integral test's equivalence (step 2):

\sum n^{-p}

converges iff

p>1

— exactly the p-series test stated above, inherited directly from Chapter 08's already-proven p-integral result.

❌ Applying the integral test to a non-decreasing function

Attempting to test $\displaystyle\sum_{n=1}^{\infty}\frac{n}{n^2+1}\cdot(-1)^{\lfloor n/2\rfloor}$ -style sign-changing terms directly with the integral test (details of the sign pattern aside) — or more simply, misapplying the test to $a_n=\sin^2(n)+1$ , which does not decrease monotonically as $n$ grows (it oscillates).

Why it breaks: the integral test's hypothesis explicitly requires $f$ to be decreasing (in addition to positive and continuous) for every $x$ beyond the starting point. Without that, the rectangle-sandwiching argument in the note above breaks down — the rectangles are no longer guaranteed to lie entirely above or entirely below the curve, so the integral and the series need not converge or diverge together.

Consequence: any series with non-monotonic or sign-changing terms needs a different test — the comparison test (Section 01), the alternating series test (Section 04), or the ratio test (Chapter 10) — never the integral test, which is only valid for positive, continuous, eventually-decreasing terms.

03 · The Limit Comparison Test

Direct term-by-term comparison (Section 01) requires finding an inequality that holds for every $n$ — sometimes awkward to set up. The limit comparison test replaces that inequality with a single limit computation.

Definition — Limit Comparison Test

Suppose $a_n, b_n > 0$ for all $n$ , and $\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n} = L$ where $0<L<\infty$ (a finite, positive number). Then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

Step-by-step — Determine convergence of $\displaystyle\sum_{n=1}^{\infty}\frac{3n^2+1}{n^4+2n}$

Identify the dominant behavior for large $n$ : the numerator behaves like

3n^2

and the denominator like

n^4

for large

n

, so

a_n

behaves like

\dfrac{3n^2}{n^4}=\dfrac{3}{n^2}

**Choose

b_n = \dfrac{1}{n^2}

, a p-series with

p=2>1

(Section 02: converges).

Compute the limit comparison:

\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{(3n^2+1)/(n^4+2n)}{1/n^2} = \lim_{n\to\infty}\frac{n^2(3n^2+1)}{n^4+2n} = \lim_{n\to\infty}\frac{3n^4+n^2}{n^4+2n}

Divide numerator and denominator by $n^4$ :

\dfrac{3+1/n^2}{1+2/n^3} \to \dfrac{3+0}{1+0}=3

n\to\infty

Conclude:

L=3

, finite and positive. Since

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

converges (step 2), the limit comparison test gives that

\displaystyle\sum\frac{3n^2+1}{n^4+2n}

converges too.

04 · The Alternating Series Test

Every test so far required non-negative terms. Series whose terms alternate in sign behave differently — the sign changes themselves can produce convergence.

Definition — Alternating Series Test

For $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}b_n$ with $b_n>0$ : if (1) $b_{n+1}\leq b_n$ for all $n$ (terms shrink, or stay equal) and (2) $\displaystyle\lim_{n\to\infty}b_n=0$ , then the series converges.

Example — The alternating harmonic series converges

$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} = 1-\frac12+\frac13-\frac14+\cdots$ . Here $b_n=\dfrac1n$ : it is decreasing ( $b_{n+1}=\frac{1}{n+1}<\frac1n=b_n$ ) and $b_n\to0$ (Chapter 09, Section 01's example). Both alternating series test conditions hold, so this series converges.

This is the single most important contrast in the whole topic: Chapter 09, Section 05 proved the plain harmonic series $\sum\frac1n$ diverges — but inserting alternating signs on the exact same terms converges. The sign pattern itself is doing real work.

❌ Dropping the 'decreasing' condition of the alternating series test

Consider $\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}b_n$ where $b_n = \dfrac{2+(-1)^n}{n}$ — so $b_n$ takes the values $\frac1n$ and $\frac3n$ alternately, and $b_n\to0$ is still true, but $b_n$ does not decrease monotonically (it bounces between roughly $\frac1n$ and $\frac3n$ instead of shrinking every single step).

Why it breaks: the alternating series test's proof (not derived in full here) relies on the partial sums oscillating with strictly shrinking amplitude — if $b_n$ fails to decrease, that oscillating-and-shrinking structure is not guaranteed, and the test's conclusion no longer follows from its hypotheses.

Consequence: $b_n\to0$ alone is not sufficient — both conditions of the alternating series test must hold together. (This particular series happens to still converge by a more delicate argument, but the alternating series test itself does not certify it, since one of its two hypotheses fails; a test only proves what its hypotheses actually guarantee.)

05 · Absolute vs. Conditional Convergence

Section 04 showed that a series can converge only because of its alternating signs. This motivates classifying series with mixed-sign terms by a stronger question: does the series still converge if every term is made positive?

Definition — Absolute and Conditional Convergence

$\sum a_n$ converges absolutely if $\sum\lvert a_n\rvert$ converges (a stronger condition — this is Chapter 10's "converges absolutely" language, made precise here).

$\sum a_n$ converges conditionally if $\sum a_n$ converges but $\sum \lvert a_n\rvert$ diverges.

Absolute convergence implies convergence — if $\sum\lvert a_n\rvert$ converges, then $\sum a_n$ converges too (proof omitted, but consistent with the comparison test: $-\lvert a_n\rvert \leq a_n \leq \lvert a_n\rvert$ sandwiches $a_n$ between two pieces built from the convergent $\sum\lvert a_n\rvert$ ).

Example — Classifying the alternating harmonic series

$\displaystyle\sum\frac{(-1)^{n-1}}{n}$ converges (Section 04). Taking absolute values: $\sum\left\lvert\frac{(-1)^{n-1}}{n}\right\rvert = \sum\frac1n$ , the plain harmonic series, which diverges (Chapter 09, Section 05).

The original series converges, but the absolute-value version diverges — so $\sum\frac{(-1)^{n-1}}{n}$ converges conditionally, not absolutely.

06 · Choosing a Test

Definition — Test Selection Checklist

Do the terms fail to approach $0$ ? → test for divergence (Chapter 09) settles it immediately.
Is it a geometric series or p-series? → apply the known formula/threshold directly (Chapter 09, Section 02 above).
Do the terms resemble a known series (geometric or p-series) after ignoring lower-order behavior? → comparison test or limit comparison test.
Are the terms $f(n)$ for a positive, continuous, decreasing $f$ , with no elementary antiderivative comparison available? → integral test.
Do the signs strictly alternate? → alternating series test, then classify absolute vs. conditional.
Does the term involve a factorial, a fixed base raised to the $n$ -th power, or resemble a power series? → ratio test (Chapter 10).

07 · Exercises

EXERCISE 12.1

Bound the given series' terms above (or below) by a p-series or geometric series whose convergence is already known.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^3+1}$ : for all $n\geq1$ , $n^3+1>n^3$ , so $\dfrac{1}{n^3+1}<\dfrac{1}{n^3}$ . $\sum\frac{1}{n^3}$ is a p-series with $p=3>1$ , converges (Section 02). By the comparison test, $\sum\frac{1}{n^3+1}$ converges too.

Use the comparison test to determine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^3+1}$ converges.

EXERCISE 12.2

Identify the dominant terms in the numerator and denominator for large $n$ , and choose a comparison series matching that behavior.

$\displaystyle\sum_{n=1}^{\infty}\frac{5n+2}{n^3-1}$ behaves like $\frac{5n}{n^3}=\frac5{n^2}$ for large $n$ . Choose $b_n=\frac1{n^2}$ (p-series, $p=2>1$ , converges). $\lim_{n\to\infty}\dfrac{a_n}{b_n} = \lim\dfrac{n^2(5n+2)}{n^3-1} = \lim\dfrac{5n^3+2n^2}{n^3-1} = 5$ (dividing by $n^3$ ). $L=5$ , finite and positive, so by the limit comparison test, $\sum\frac{5n+2}{n^3-1}$ converges.

Use the limit comparison test to determine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{5n+2}{n^3-1}$ converges.

EXERCISE 12.3

Check that the integral test's three hypotheses (positive, continuous, decreasing) hold, then evaluate the corresponding improper integral using Chapter 08's tools.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(\ln n)^2}$ (for $n\geq2$ , since $\ln1=0$ ): let $f(x)=\dfrac{1}{x(\ln x)^2}$ , positive, continuous, and decreasing for $x\geq2$ . $\int_2^\infty \dfrac{dx}{x(\ln x)^2}$ : substitute $u=\ln x$ , $du=\dfrac{dx}{x}$ (Chapter 07 u-substitution), giving $\int \dfrac{du}{u^2} = -\dfrac1u+C = -\dfrac{1}{\ln x}+C$ . Evaluate: $\lim_{t\to\infty}\left[-\dfrac{1}{\ln x}\right]_2^t = \lim_{t\to\infty}\left(-\dfrac{1}{\ln t}+\dfrac{1}{\ln2}\right) = 0+\dfrac{1}{\ln2} = \dfrac{1}{\ln2}$ , finite. The integral converges, so by the integral test, the series converges too.

Use the integral test to determine whether $\displaystyle\sum_{n=2}^{\infty}\dfrac{1}{n(\ln n)^2}$ converges.

EXERCISE 12.4

Check both alternating series test conditions (decreasing terms, terms shrinking to zero), then separately check convergence of the absolute-value series.

$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^2}$ : $b_n=\frac1{n^2}$ is decreasing and $\to0$ , so by the alternating series test the series converges. Absolute values: $\sum\frac{1}{n^2}$ , a p-series with $p=2>1$ , which converges. Since the absolute-value series also converges, the original series converges absolutely, not merely conditionally.

Show that $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n-1}}{n^2}$ converges, and determine whether it converges absolutely or only conditionally.

EXERCISE 12.5

The presence of a factorial in the terms is the strongest signal to reach for the ratio test rather than a comparison-style test.

$\displaystyle\sum_{n=1}^{\infty}\frac{n!}{2^n}$ : ratio test, $\left\lvert\dfrac{a_{n+1}}{a_n}\right\rvert = \dfrac{(n+1)!}{2^{n+1}}\cdot\dfrac{2^n}{n!} = \dfrac{n+1}{2} \to \infty$ as $n\to\infty$ . Since $\rho=\infty>1$ , the series diverges by the ratio test (Chapter 10) — factorial growth in the numerator overwhelms the exponential denominator.

Use the ratio test to determine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{n!}{2^n}$ converges.

EXERCISE 12.6

A stream of alternating gains and losses, shrinking in size and tending to zero, is exactly the setup for the alternating series test.

A trading strategy's realized monthly P&L follows $p_n = (-1)^{n-1}\dfrac{1000}{n}$ dollars in month $n$ (alternating gains and losses, shrinking in size). Total realized P&L across all months: $\sum_{n=1}^\infty (-1)^{n-1}\dfrac{1000}{n} = 1000\sum(-1)^{n-1}\dfrac1n$ . Since $b_n=\frac1n$ is decreasing and $\to0$ , the alternating series test gives convergence — the infinite stream of P&L sums to a finite total, even though the un-signed magnitudes $\sum\frac{1000}{n}$ (the harmonic series scaled by 1000) would diverge. This is a conditionally convergent total: it depends on the exact alternating order of gains and losses, not just their sizes.

A trading strategy's monthly P&L is modeled by $p_n = (-1)^{n-1}\dfrac{1000}{n}$ dollars in month $n$ . Determine whether the total P&L over infinitely many months converges, and state whether it does so absolutely or conditionally.

08 · Chapter Summary

Concept	Formula / Rule
Comparison test	$0\leq a_n\leq b_n$ : $b_n$ converges $\Rightarrow a_n$ converges; $a_n$ diverges $\Rightarrow b_n$ diverges
p-series	$\sum 1/n^p$ converges iff $p>1$
Integral test	$a_n=f(n)$ , $f$ positive/continuous/decreasing: series and $\int_1^\infty f\,dx$ share convergence
Limit comparison test	$\lim a_n/b_n = L\in(0,\infty)$ : both converge or both diverge
Alternating series test	$b_n$ decreasing, $b_n\to0$ $\Rightarrow \sum(-1)^{n-1}b_n$ converges
Absolute convergence	$\sum\lvert a_n\rvert$ converges $\Rightarrow \sum a_n$ converges
Conditional convergence	$\sum a_n$ converges but $\sum\lvert a_n\rvert$ diverges
Ratio test	Recap from Chapter 10 — best for factorials, fixed bases to the $n$

Next: Chapter 13 — Parametric Equations and Polar Coordinates, which describes curves that cannot be written as a single function $y=f(x)$ .