Chapter 22 — Convergence Tests

22.1 The Hardest Chapter, and Why It Is Worth It

Back in Chapter 1 we warned you that this would be the hardest single chapter in the book. Here is why — and here is the promise that makes it survivable.

For almost every infinite series you will ever meet, you cannot compute the exact sum. The harmonic series $\sum 1/n$ has no closed form. The series $\sum 1/n^3$ has no closed form (its value, Apéry's constant, was only proven irrational in 1978). Yet for a vast range of series you can answer the one question that matters most: does it converge or diverge? Does adding infinitely many terms produce a finite number, or does the sum run away to infinity?

The difficulty is not that any single test is hard. Each of the seven tests in this chapter is a one- or two-line procedure. The difficulty is choosing — given a brand-new series, which test do you reach for? That is a recognition skill, like a doctor reading a chart, and the entire chapter is engineered to teach it. We will end every major idea by folding it back into a single decision framework (§22.10) that turns "which test?" from a mystery into a flowchart.

The Key Insight. You do not need to memorize seven unrelated tricks. You need to recognize structure. Factorials and exponentials cry out for the ratio test; an $n$-th power cries out for the root test; terms that decay like $1/n^p$ cry out for limit comparison; alternating signs cry out for the alternating series test. Learn to read the shape of a series and the right test announces itself.

Here is the full toolbox, in the order we will build it:

1. Divergence test (from Chapter 21): if $a_n \not\to 0$, the series diverges. Always your first glance.
2. Integral test: for positive decreasing $f$ with $a_n = f(n)$, the series and the improper integral $\int_1^\infty f$ (Chapter 17) live or die together. This gives the $p$-series rule.
3. Direct comparison test: trap your series between known series.
4. Limit comparison test: match your series to one whose behavior you know by comparing leading-order decay.
5. Ratio test: built for factorials and exponentials.
6. Root test: built for $n$-th powers.
7. Alternating series test: for series whose signs flip, with a built-in error bound.

Layered on top is the distinction between absolute and conditional convergence — and the genuinely shocking Riemann rearrangement theorem, which says that for a conditionally convergent series, the order of the terms determines the sum.

Two recurring themes from this book run straight through the chapter. Approximation is the soul of calculus: every convergent series is an infinite approximation, and the alternating-series error bound tells you exactly how good a finite truncation is. And hand computation builds understanding while machine computation builds power: you will apply each test by hand, then watch Python confirm the convergence rate numerically.

22.2 The Integral Test

We begin with the test that ties this chapter back to integration — and in doing so settles the single most-cited convergence question in all of calculus.

Integral Test. Suppose $f$ is positive, continuous, and decreasing on $[1, \infty)$, and let $a_n = f(n)$. Then $$\sum_{n=1}^\infty a_n \quad\text{and}\quad \int_1^\infty f(x)\,dx \quad\text{converge or diverge together.}$$ (If they converge, they generally converge to different values — the test compares only whether, not to what.)

Geometric Intuition. Draw the curve $y = f(x)$ and stack rectangles of width $1$ under it, one per integer. Using left endpoints, each rectangle has height $f(n)$ and sits above the curve over $[n, n+1]$ — so the rectangles' total area $\sum a_n$ overshoots the integral. Using right endpoints, the rectangles sit below the curve, and their area undershoots it. Sandwiching the series between two copies of the same integral (shifted by one term) forces the series and integral to have the same fate. Picture this — it is the proof.

Sketch of the proof. Because $f$ decreases, on the interval $[n, n+1]$ we have $f(n+1) \le f(x) \le f(n)$. Integrate across $[n,n+1]$: $$a_{n+1} = f(n+1) \le \int_n^{n+1} f(x)\,dx \le f(n) = a_n.$$ Summing from $n=1$ to $N-1$ traps the partial integral between two partial sums: $$\sum_{n=2}^{N} a_n \le \int_1^{N} f(x)\,dx \le \sum_{n=1}^{N-1} a_n.$$ If the integral converges as $N \to \infty$, the left inequality bounds the partial sums above; a bounded increasing sequence of partial sums converges (Chapter 20). If the integral diverges, the right inequality forces the partial sums to infinity. Either way, series and integral share the verdict. $\blacksquare$

22.2.1 The $p$-Series Rule

The classic payoff is the family $\sum 1/n^p$. Compare it to $\int_1^\infty x^{-p}\,dx$, which we evaluated as an improper integral in Chapter 17:

$$\int_1^\infty \frac{dx}{x^p} = \begin{cases} \dfrac{1}{p-1} & p > 1 \quad(\text{converges}) \\[4pt] \infty & p \le 1 \quad(\text{diverges}). \end{cases}$$

Therefore:

The $p$-Series Rule. $\displaystyle\sum_{n=1}^\infty \frac{1}{n^p}$ converges if and only if $p > 1$.

This one line is worth committing to memory; you will lean on it constantly, especially as the comparison yardstick in §22.4. The borderline case $p = 1$ is the harmonic series $\sum 1/n$, which diverges — barely, but unmistakably — exactly as the integral $\int_1^\infty dx/x = \ln x \big|_1^\infty = \infty$ does. The divergence is glacial: to push the partial sum past $20$, you need more than $250$ million terms, because $\sum_{n=1}^N 1/n \approx \ln N$. It diverges, but it is in no hurry.

Historical Note. The harmonic series' divergence was first proved around 1350 by the French scholar Nicole Oresme, using a grouping argument that needs no calculus at all: $\tfrac13+\tfrac14 > \tfrac12$, then $\tfrac15+\cdots+\tfrac18 > \tfrac12$, then the next eight terms again exceed $\tfrac12$, and so on forever — infinitely many half-units, so the sum is infinite. The convergent companion question — the exact value of $\sum 1/n^2$ — resisted the best mathematicians of Europe for ninety years (the "Basel problem") until Leonhard Euler stunned everyone in 1734 by proving it equals $\pi^2/6$. That a sum of reciprocal squares should conjure $\pi$ remains one of the great surprises of mathematics.

Check Your Understanding. Does $\sum_{n=1}^\infty n^{-3/2}$ converge? What about $\sum_{n=1}^\infty n^{-2/3}$?

Answer

For $n^{-3/2}$, $p = 3/2 > 1$, so it converges. For $n^{-2/3}$, $p = 2/3 \le 1$, so it diverges. The exponent crossing $1$ is the entire story.

22.2.2 Worked Examples — Logarithmic Borderlines

The integral test shines when the integrand has a clean antiderivative via substitution. The next two examples are nearly identical in appearance yet land on opposite sides of convergence — a vivid demonstration of how delicate the boundary is.

Worked Example 22.2.A. Test $\displaystyle\sum_{n=2}^\infty \frac{1}{n \ln n}$.

The function $f(x) = 1/(x\ln x)$ is positive, continuous, and decreasing on $[2,\infty)$. Substitute $u = \ln x$, $du = dx/x$: $$\int_2^\infty \frac{dx}{x \ln x} = \int_{\ln 2}^\infty \frac{du}{u} = \ln u \Big|_{\ln 2}^\infty = \infty.$$ The integral diverges, so the series diverges.

Worked Example 22.2.B. Test $\displaystyle\sum_{n=2}^\infty \frac{1}{n (\ln n)^2}$.

Same substitution $u = \ln x$: $$\int_2^\infty \frac{dx}{x (\ln x)^2} = \int_{\ln 2}^\infty \frac{du}{u^2} = -\frac{1}{u}\Big|_{\ln 2}^\infty = \frac{1}{\ln 2}.$$ The integral converges (to $1/\ln 2 \approx 1.443$), so the series converges.

These two series differ only by one extra power of $\ln n$ in the denominator, yet one diverges and the other converges. No comparison or ratio test sees this distinction easily — the integral test makes it transparent. (And note: the series converges, but not to $1/\ln 2$; that is the integral's value, not the sum's. The test reports only the verdict.)

Warning. The integral test demands that $f$ be decreasing — at least eventually. If $f$ wiggles up and down, $a_n = f(n)$ need not track the integral at all, and the sandwich argument collapses. Before invoking the test, confirm monotonicity, usually by checking $f'(x) \le 0$ for large $x$. Continuity and positivity matter too, but the decreasing condition is the one students forget.

22.3 The Direct Comparison Test

The simplest idea in the chapter: a positive series cannot be larger than something finite and still escape to infinity.

Direct Comparison Test. Suppose $0 \le a_n \le b_n$ for all $n$ (or for all large $n$). Then: - if $\sum b_n$ converges, so does $\sum a_n$ (the smaller series is dominated); - if $\sum a_n$ diverges, so does $\sum b_n$ (the larger series is dragged up).

Geometric Intuition. Stack the terms as bars. If the tall bars $b_n$ have finite total height and your bars $a_n$ are each shorter, your total is finite too. If the short bars $a_n$ already add to infinity, the taller bars certainly do. The logic only runs "downward for convergence, upward for divergence" — getting the direction backward is the classic trap below.

Worked Example 22.3.A. Test $\displaystyle\sum_{n=1}^\infty \frac{1}{n^2 + 1}$.

Since $n^2 + 1 > n^2$, we have $\dfrac{1}{n^2+1} < \dfrac{1}{n^2}$. The comparison series $\sum 1/n^2$ converges ($p = 2 > 1$), so by direct comparison our smaller series converges.

Worked Example 22.3.B. Test $\displaystyle\sum_{n=1}^\infty \frac{1}{2^n - 1}$.

For $n \ge 1$, $2^n - 1 \ge 2^{n-1}$ (since $2^n - 1 \ge 2^n/2$ when $2^n \ge 2$), so $\dfrac{1}{2^n - 1} \le \dfrac{2}{2^n} = 2\cdot\left(\tfrac12\right)^n$. The geometric series $\sum 2\cdot(1/2)^n$ converges (Chapter 21, $|r| = 1/2 < 1$), so our series converges.

Worked Example 22.3.C (a divergence by comparison). Test $\displaystyle\sum_{n=1}^\infty \frac{\ln n}{n}$.

For $n \ge 3$, $\ln n > 1$, so $\dfrac{\ln n}{n} > \dfrac{1}{n}$. We have bounded our series below by the divergent harmonic series $\sum 1/n$. By direct comparison the larger series diverges. Note the discipline this requires: to prove divergence we needed a smaller divergent series underneath — the mirror image of Example 22.3.A, where proving convergence needed a larger convergent series on top.

Common Pitfall. Many students try to prove a series converges by comparing it to something larger that diverges — which proves nothing. To show convergence you must bound your series above by a convergent series; to show divergence you must bound it below by a divergent series. Comparing $\frac{1}{n^2+1}$ to the divergent $\frac{1}{n}$ (which is larger) tells you absolutely nothing. Match the direction of the inequality to the direction of the conclusion.

When Direct Comparison Gets Awkward

Direct comparison needs an honest inequality, and sometimes the inequality runs the wrong way for small $n$. Consider $\sum \frac{2n+1}{n^3 + 5}$. Its leading behavior is $\frac{2n}{n^3} = \frac{2}{n^2}$, so we expect convergence. But the clean inequality $\frac{2n+1}{n^3+5} \le \frac{2}{n^2}$ is not obvious to verify, and for some early $n$ the term can exceed simple candidates. Rather than fight the algebra, we reach for a tool that compares asymptotic behavior and ignores the messy constants entirely: the limit comparison test.

22.4 The Limit Comparison Test

This is, in practice, the most-used comparison tool — because it asks only how two series behave for large $n$, not for an exact inequality at every term.

Limit Comparison Test. Let $a_n, b_n > 0$ and suppose $\displaystyle\lim_{n\to\infty} \frac{a_n}{b_n} = c$. - If $0 < c < \infty$: both series converge or both diverge (they share a fate). - If $c = 0$ and $\sum b_n$ converges, then $\sum a_n$ converges. - If $c = \infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.

The first case is the workhorse. When $a_n/b_n$ tends to a positive finite limit, the two series are "asymptotically proportional" — eventually $a_n$ is sandwiched between $\tfrac{c}{2} b_n$ and $2c\, b_n$, so direct comparison applies in both directions at once. The strategy is almost mechanical: find the leading-order behavior of $a_n$, set $b_n$ equal to it, and take the limit.

Worked Example 22.4.A. Test $\displaystyle\sum_{n=1}^\infty \frac{2n+1}{n^3 + 5}$ (the awkward one from above).

The leading behavior is $\frac{2n}{n^3} = \frac{2}{n^2}$, so compare with $b_n = 1/n^2$: $$\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{2n+1}{n^3+5}\cdot n^2 = \lim_{n\to\infty} \frac{2n^3 + n^2}{n^3 + 5} = 2.$$ The limit $c = 2$ is finite and positive, and $\sum 1/n^2$ converges, so our series converges — no painful inequality required.

Worked Example 22.4.B. Test $\displaystyle\sum_{n=1}^\infty \frac{1}{\sqrt{n^4 + 1}}$.

For large $n$, $\sqrt{n^4+1} \approx n^2$, so compare with $b_n = 1/n^2$: $$\lim_{n\to\infty} \frac{n^2}{\sqrt{n^4 + 1}} = \lim_{n\to\infty} \frac{1}{\sqrt{1 + 1/n^4}} = 1.$$ Positive finite limit, and $\sum 1/n^2$ converges, so the series converges.

Worked Example 22.4.C (a divergent case). Test $\displaystyle\sum_{n=1}^\infty \frac{n+4}{\sqrt{n^3 + n}}$.

Leading behavior: $\frac{n}{\sqrt{n^3}} = \frac{n}{n^{3/2}} = \frac{1}{n^{1/2}}$, so compare with $b_n = n^{-1/2}$: $$\lim_{n\to\infty} \frac{(n+4)/\sqrt{n^3+n}}{1/\sqrt n} = \lim_{n\to\infty} \frac{(n+4)\sqrt n}{\sqrt{n^3 + n}} = \lim_{n\to\infty}\frac{n^{3/2}+4n^{1/2}}{n^{3/2}\sqrt{1 + 1/n^2}} = 1.$$ Positive finite limit, and $\sum n^{-1/2}$ diverges ($p = 1/2 \le 1$), so the series diverges.

Check Your Understanding. Which $b_n$ would you pick to test $\sum \frac{3n^2 - 1}{n^4 + 2n}$ by limit comparison, and what is the verdict?

Answer

Leading behavior is $\frac{3n^2}{n^4} = \frac{3}{n^2}$, so take $b_n = 1/n^2$. Then $a_n/b_n \to 3$, a positive finite limit. Since $\sum 1/n^2$ converges, the series converges.

Real-World Application — Convergence of network and queueing sums (engineering / computer science). The expected number of items waiting in many queueing models takes the form $\sum_n n \cdot P(N = n)$, where $P(N=n)$ is the probability of $n$ items in the system. Whether the mean queue length is finite — the difference between a stable system and one that grows without bound — is exactly a series-convergence question. For heavy-tailed arrival distributions where $P(N=n)\sim C/n^{p}$, limit comparison against a $p$-series instantly tells the engineer whether the average backlog is finite (mean exists when $p > 2$) or infinite. The same logic governs whether the variance of a power-law distribution exists, a question that decides whether standard statistical tools even apply.

22.5 The Ratio Test

When a series contains factorials, exponentials, or products that grow multiplicatively, the natural thing to study is the ratio of consecutive terms.

Ratio Test. For a series $\sum a_n$, let $$\rho = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|.$$ - If $\rho < 1$: the series converges absolutely. - If $\rho > 1$ (or $\rho = \infty$): the series diverges. - If $\rho = 1$: the test is inconclusive — use something else.

Why it works. If eventually $|a_{n+1}| \le r\,|a_n|$ for some fixed $r < 1$, then the tail of the series is dominated, term by term, by a geometric series with ratio $r$ — which converges. The ratio $\rho$ measures the "local multiplication factor"; if each term is a definite fraction of the one before, the sum is geometric-like and finite. If the terms keep growing ($\rho > 1$), they cannot even tend to zero, so the series diverges by the divergence test.

Worked Example 22.5.A. Test $\displaystyle\sum_{n=1}^\infty \frac{1}{n!}$.

$$\frac{a_{n+1}}{a_n} = \frac{1/(n+1)!}{1/n!} = \frac{n!}{(n+1)!} = \frac{1}{n+1} \to 0 = \rho < 1.$$ The series converges absolutely. (Including the $n = 0$ term, $\sum_{n=0}^\infty 1/n! = e$, the defining series for Euler's number — an anchor we will meet again in Chapter 24.)

Worked Example 22.5.B. Test $\displaystyle\sum_{n=1}^\infty \frac{n^2}{2^n}$.

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{2^{n+1}}\cdot\frac{2^n}{n^2} = \frac{(n+1)^2}{2 n^2} = \frac{1}{2}\left(1 + \frac1n\right)^2 \to \frac{1}{2} = \rho < 1.$$ Converges. The exponential $2^n$ in the denominator overwhelms the polynomial $n^2$ — a fact the ratio test exposes instantly.

Worked Example 22.5.C (divergence). Test $\displaystyle\sum_{n=1}^\infty \frac{n^n}{n!}$.

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^n} = \frac{(n+1)^{n+1}}{(n+1)\,n^n} = \frac{(n+1)^n}{n^n} = \left(1 + \frac1n\right)^n \to e \approx 2.718 > 1.$$ Since $\rho = e > 1$, the series diverges. (This limit $(1+1/n)^n \to e$ is the famous definition of $e$ from Chapter 11 — a satisfying place to see it resurface.)

Computational Note. The ratio test's threshold $\rho < 1$ is the same condition as for a geometric series, and that is no coincidence: the ratio test is "local geometric comparison." Numerically, $\rho$ is the asymptotic per-term shrink factor, so a series with $\rho = 0.5$ loses about one bit of magnitude per term and converges fast; a series with $\rho = 0.99$ converges agonizingly slowly. We will see this rate difference in the Python block of §22.9.

When the Ratio Test Fails

The fatal weakness: for any $p$-series, $\rho = 1$. $$\frac{a_{n+1}}{a_n} = \frac{1/(n+1)^p}{1/n^p} = \left(\frac{n}{n+1}\right)^p \to 1.$$ So the ratio test cannot distinguish the convergent $\sum 1/n^2$ from the divergent $\sum 1/n$ — both give $\rho = 1$. Whenever the terms decay only like a power of $n$, the ratio test is silent; reach for the integral test or limit comparison instead. The ratio test is a tool for multiplicative structure (factorials, exponentials), not power-law structure.

22.6 The Root Test

A close cousin of the ratio test, tuned for terms that are $n$-th powers.

Root Test. For a series $\sum a_n$, let $$\rho = \lim_{n\to\infty} |a_n|^{1/n}.$$ - If $\rho < 1$: converges absolutely. - If $\rho > 1$ (or $\rho = \infty$): diverges. - If $\rho = 1$: inconclusive.

The outcomes mirror the ratio test exactly, and for good reason: both compare the series to a geometric one. The root test is the right tool when the whole term is raised to the $n$-th power, because then $|a_n|^{1/n}$ peels the exponent off cleanly.

Worked Example 22.6.A. Test $\displaystyle\sum_{n=1}^\infty \left(\frac{n}{n+1}\right)^{n^2}$.

$$|a_n|^{1/n} = \left(\frac{n}{n+1}\right)^{n^2/n} = \left(\frac{n}{n+1}\right)^{n} = \left(1 + \frac1n\right)^{-n} \to e^{-1} = \frac{1}{e} \approx 0.368 < 1.$$ The series converges. The exponent $n^2$ would make the ratio test a nightmare; the root test handles it in one line.

Worked Example 22.6.B. Test $\displaystyle\sum_{n=2}^\infty \frac{1}{(\ln n)^n}$.

$$|a_n|^{1/n} = \frac{1}{\ln n} \to 0 < 1.$$ Converges — and quite fast, since $\ln n \to \infty$ makes the base shrink without bound.

Worked Example 22.6.C. Test $\displaystyle\sum_{n=1}^\infty \left(\frac{2n}{n+3}\right)^{n}$.

$$|a_n|^{1/n} = \frac{2n}{n+3} \to 2 > 1.$$ The series diverges: the terms do not even approach zero.

Math Major Sidebar — Root test is strictly stronger than ratio test. Whenever the ratio limit exists, the root limit exists and equals it, so anything the ratio test decides the root test decides too. But the root test (stated with $\limsup$) can succeed where the ratio limit fails to exist. Consider $a_n = 2^{-n}$ for even $n$ and $a_n = 2^{-n+1}$ for odd $n$. The consecutive ratios oscillate between $1/4$ and $1$, so $\lim |a_{n+1}/a_n|$ does not exist and the ratio test is helpless. But $|a_n|^{1/n} \to 1/2 < 1$, so the root test cleanly proves convergence. In real analysis this is why the radius of convergence of a power series (Chapter 23) is defined via the root test (the Cauchy–Hadamard formula), not the ratio test.

22.7 The Alternating Series Test and Its Error Bound

Every test so far assumed positive terms. Series whose signs flip — like $1 - \tfrac12 + \tfrac13 - \tfrac14 + \cdots$ — get a dedicated tool, and it comes with a bonus the others lack: a guaranteed error bound.

Alternating Series Test (Leibniz's Test). Consider $\displaystyle\sum_{n=1}^\infty (-1)^{n+1} b_n$ with $b_n > 0$. If 1. $b_n$ is eventually decreasing ($b_{n+1} \le b_n$), and 2. $b_n \to 0$, then the series converges.

Geometric Intuition. Plot the partial sums $S_1, S_2, S_3, \dots$ on a number line. Each new term jumps you forward then backward by a shrinking amount: $S_1$ overshoots, $S_2$ undershoots, $S_3$ overshoots less, and so on. The partial sums oscillate inward like a coin spiraling down a funnel, the odd ones above the limit and the even ones below, the gap between them ($b_{n+1}$) closing to zero. They are squeezed onto a single limit point $L$. That picture is the whole proof — and it instantly delivers the error bound below.

Worked Example 22.7.A (the alternating harmonic series). Test $\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac12 + \frac13 - \frac14 + \cdots$.

Here $b_n = 1/n$ is decreasing and $\to 0$, so by the alternating series test it converges. (Its sum is $\ln 2$; we derive this from a Taylor series in Chapter 23.) The remarkable part: the non-alternating version $\sum 1/n$ diverges. Flipping signs rescues convergence — the cancellation between adjacent terms is doing real work.

Worked Example 22.7.B. Test $\displaystyle\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt n}$.

$b_n = 1/\sqrt n$ is decreasing and $\to 0$, so the series converges. Again, the absolute version $\sum 1/\sqrt n$ diverges ($p = 1/2$) — a theme we formalize as conditional convergence in §22.8.

The Alternating Series Remainder

The funnel picture has a precise consequence. If the series converges to $L$ and you stop at the $N$-th partial sum $S_N$, the true value $L$ lies between $S_N$ and $S_{N+1}$, so your error is no larger than the size of the very next term:

Alternating Series Error Bound. For a convergent alternating series satisfying Leibniz's conditions, $$|L - S_N| \le b_{N+1}.$$ The error of a partial sum is at most the first omitted term — and it has the same sign as that term.

This is extraordinarily useful: it turns a divergent-looking approximation into a certified one. To approximate $L$ to within a tolerance $\varepsilon$, just sum until the next term drops below $\varepsilon$.

Worked Example 22.7.C (certified approximation). How many terms of $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \ln 2$ guarantee an error below $0.01$?

We need $b_{N+1} = \frac{1}{N+1} \le 0.01$, i.e. $N + 1 \ge 100$, so $N \ge 99$. Summing the first $99$ terms guarantees $|{\ln 2} - S_{99}| \le 0.01$. (This also reveals how slowly the alternating harmonic series converges — error shrinks only like $1/N$. Contrast a series with $b_n = 1/n!$, where a handful of terms nails many digits.)

Common Pitfall. The error bound $|L - S_N| \le b_{N+1}$ applies only to series that satisfy both Leibniz conditions — alternating signs and terms decreasing to zero. Students sometimes slap "the error is the next term" onto a non-alternating series, or onto an alternating series whose terms are not yet monotonically decreasing. Neither is valid. Verify decreasing-to-zero before trusting the bound.

Real-World Application — Computing $\pi$ and certified numerics (computer science / numerical analysis). The Leibniz series $\frac{\pi}{4} = 1 - \frac13 + \frac15 - \frac17 + \cdots$ is alternating with $b_n = 1/(2n-1)$, so the error bound tells you exactly how many terms guarantee a target precision — no guesswork. This "the error is the next term" guarantee is the foundation of certified numerical computing: when a result must come with a provable error bar (in scientific software, control systems, or interval arithmetic), alternating-series tails are prized precisely because their truncation error is bounded by a single computable number.

22.8 Absolute vs. Conditional Convergence

We can now name the phenomenon that has appeared three times: some series converge because of sign cancellation, and fall apart without it.

Definitions. A series $\sum a_n$ is absolutely convergent if $\sum |a_n|$ converges. It is conditionally convergent if it converges but $\sum |a_n|$ diverges.

The fundamental fact linking them:

Absolute Convergence Theorem. If $\sum |a_n|$ converges, then $\sum a_n$ converges. (Absolute convergence implies convergence.)

So "absolutely convergent" is the stronger condition. The reason this matters strategically: it lets you apply the positive-term tests (integral, comparison, ratio, root) to a series with mixed signs — just test $\sum |a_n|$. If that converges, you are done, and you get a more robust kind of convergence for free.

Examples.

• $\displaystyle\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$: the absolute series $\sum 1/n^2$ converges, so this is absolutely convergent.
• $\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$: converges (alternating series test) but $\sum 1/n$ diverges, so this is conditionally convergent.
• $\displaystyle\sum_{n=1}^\infty \frac{\sin n}{n^2}$: since $|\sin n|/n^2 \le 1/n^2$, direct comparison shows $\sum |a_n|$ converges, so this is absolutely convergent — even though the signs of $\sin n$ follow no simple pattern.

Check Your Understanding. Classify $\sum_{n=1}^\infty \frac{(-1)^n}{n^{3/2}}$ and $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt[3]{n}}$ as absolutely convergent, conditionally convergent, or divergent.

Answer

First: $\sum 1/n^{3/2}$ converges ($p = 3/2 > 1$), so the series is absolutely convergent. Second: $\sum 1/n^{1/3}$ diverges ($p = 1/3 \le 1$), but the alternating series converges since $1/\sqrt[3]{n}\to 0$ decreasing — so it is conditionally convergent.

Math Major Sidebar — The Riemann Rearrangement Theorem. Here is one of the most startling theorems in analysis. For an absolutely convergent series, you may shuffle the terms into any order at all and the sum never changes — it behaves exactly like a finite sum. But for a conditionally convergent series, the order is everything:

Riemann Rearrangement Theorem (1854). If $\sum a_n$ converges conditionally, then for any target $L \in \mathbb{R}$ — or $\pm\infty$ — there is a rearrangement of the same terms whose partial sums converge to $L$ (or diverge).

The mechanism is almost mischievous. A conditionally convergent series has positive terms summing to $+\infty$ and negative terms summing to $-\infty$, while both kinds individually shrink to zero. To hit a target $L$, add positive terms until you just exceed $L$, then add negative terms until you just dip below $L$, then more positives, and so on. Because the leftover terms keep shrinking, the overshoot and undershoot shrink too, and the partial sums home in on $L$. The alternating harmonic series sums to $\ln 2 \approx 0.693$ in its natural order — yet by reordering the exact same terms, you can make it sum to $0$, to $\pi$, or to anything you please. Infinite addition is not commutative. This is precisely why absolute convergence is the "right," order-independent notion of convergence for rigorous work, and why physicists and numerical analysts treat conditionally convergent sums with great caution.

22.9 Computing Convergence Rates in Python

Hand computation tells you whether a series converges; Python lets you see how fast. The rate of convergence is not academic — it decides how many terms a real computation needs, and it dramatizes the gulf between a ratio-test series ($\rho < 1$, geometric-fast) and a $p$-series (algebraically slow). We follow the book's three-tier pattern: pose analytically, reason by hand, then verify numerically.

# Compare how fast three convergent series approach their limits,
# and CERTIFY the alternating-harmonic error with the AST bound.
import numpy as np

def partial_sums(term, N):
    """Return the array of partial sums S_1, ..., S_N."""
    a = np.array([term(n) for n in range(1, N + 1)])
    return np.cumsum(a)

N = 50

# 1) sum 1/n^2  -> pi^2/6  (p-series, p=2: SLOW, error ~ 1/N)
p_series = partial_sums(lambda n: 1 / n**2, N)
true_p = np.pi**2 / 6
print("1/n^2  : S_50 =", round(p_series[-1], 6),
      " error =", round(true_p - p_series[-1], 6))     # error ~ 0.0198  (slow!)

# 2) sum 1/2^n  -> 1       (geometric, rho=1/2: FAST, error ~ (1/2)^N)
geom = partial_sums(lambda n: 1 / 2**n, N)
print("1/2^n  : S_50 =", round(geom[-1], 12),
      " error =", 1 - geom[-1])                          # error ~ 8.9e-16 (machine zero!)

# 3) sum (-1)^(n+1)/n -> ln 2  (alternating: error <= b_{N+1} = 1/(N+1))
alt = partial_sums(lambda n: (-1)**(n + 1) / n, N)
true_alt = np.log(2)
actual_error = abs(true_alt - alt[-1])
ast_bound = 1 / (N + 1)                                  # first omitted term
print("alt-harm: S_50 =", round(alt[-1], 6),
      " actual err =", round(actual_error, 6),
      " AST bound =", round(ast_bound, 6))
# actual err ~ 0.009902  <=  AST bound ~ 0.019608   (bound holds, as guaranteed)

The printed output tells the whole story. After $50$ terms the geometric series is accurate to machine precision ($\sim 10^{-16}$), the $p$-series is still wrong in the second decimal place, and the alternating harmonic series obeys its certified bound $|{\ln 2} - S_{50}| \le 1/51$. This is why the ratio and root tests matter so much in practice: a series with $\rho < 1$ does not merely converge — it converges fast enough to compute with.

Computational Note. Notice the third check does more than confirm convergence: it verifies the error theorem numerically. The actual error ($\approx 0.0099$) sits comfortably below the AST bound ($\approx 0.0196$), exactly as §22.7 promised, and it carries the sign of the first omitted term. When you need a sum to a guaranteed tolerance, this is the loop you write — keep adding terms until $b_{N+1}$ drops below your target $\varepsilon$, and the bound certifies the rest.

22.10 The Decision Framework: Which Test, When

This is the heart of the chapter — the part that makes everything learnable. Faced with a brand-new series $\sum a_n$, run down this checklist in order. The first matching rule usually wins.

Step 0 — Glance at the terms. Does $a_n \to 0$? - No (or unclear it goes to zero) → diverges by the divergence test. Stop. (E.g. $\sum \frac{n}{n+1}$, terms $\to 1 \ne 0$.) - Yes → continue; the divergence test is inconclusive and you need a real test.

Step 1 — Is it a recognizable named series? - Geometric $\sum r^n$ → converges iff $|r| < 1$ (Chapter 21). Stop. - $p$-series $\sum 1/n^p$ → converges iff $p > 1$. Stop. - Telescoping → sum the collapsing partial sums directly (Chapter 21). Stop.

Step 2 — Are the signs alternating? - Yes → try the alternating series test (check $b_n \downarrow 0$). If you also want absolute vs. conditional, test $\sum |a_n|$ separately with the positive-term tools below.

Step 3 — Do factorials, exponentials, or $n^n$ appear? - Yes → try the ratio test. (Best tool for multiplicative growth.)

Step 4 — Is the whole term an $n$-th power, $a_n = (\text{stuff})^n$? - Yes → try the root test.

Step 5 — Do the terms behave like $1/n^p$ for large $n$ (a ratio of polynomials or roots)? - Yes → use the limit comparison test against $1/n^p$, with $p$ = (degree of denominator) − (degree of numerator).

Step 6 — Is there a clean, decreasing integrand with an easy antiderivative (especially $\ln$-type terms)? - Yes → use the integral test.

Step 7 — Can you bound the term by a known series with an honest inequality? - Yes → use the direct comparison test (above by convergent for convergence; below by divergent for divergence).

A compact version of the same logic:

                    ┌─────────────────────────────┐
   START: ∑ aₙ ───▶ │ Does aₙ → 0 ?               │
                    └──────────────┬──────────────┘
                          no ◀─────┴─────▶ yes
                    DIVERGES                │
                  (divergence test)         ▼
                              ┌───────────────────────────────┐
                              │ Geometric / p-series /         │
                              │ telescoping?  → use formula     │
                              └───────────────┬─────────────────┘
                                   no         │
                                              ▼
                              ┌───────────────────────────────┐
                              │ Signs alternate? → ALT. SERIES │
                              │ TEST (+ test ∑|aₙ| for abs/cond)│
                              └───────────────┬─────────────────┘
                                   no         │
                                              ▼
                              ┌───────────────────────────────┐
                              │ Factorials / exponentials / nⁿ?│
                              │       → RATIO TEST              │
                              └───────────────┬─────────────────┘
                                   no         │
                                              ▼
                              ┌───────────────────────────────┐
                              │ Whole term is (…)ⁿ ?           │
                              │       → ROOT TEST               │
                              └───────────────┬─────────────────┘
                                   no         │
                                              ▼
                              ┌───────────────────────────────┐
                              │ Behaves like 1/n^p (poly ratio)?│
                              │   → LIMIT COMPARISON vs 1/n^p   │
                              └───────────────┬─────────────────┘
                                   no         │
                                              ▼
                              ┌───────────────────────────────┐
                              │ Decreasing, easy ∫ (ln-types)? │
                              │   → INTEGRAL TEST               │
                              │ else → DIRECT COMPARISON        │
                              └─────────────────────────────────┘

The Key Insight. The decision framework is the single most valuable thing in this chapter. Photocopy it, tape it above your desk, and run every series through it until the routing becomes automatic. Expert problem-solvers are not faster at any one test — they are faster at recognizing which test the series is asking for. That recognition is the skill the exam is really testing.

Worked Examples — The Framework in Action

Worked Example 22.10.A. Test $\displaystyle\sum_{n=1}^\infty \frac{n^2}{e^n}$.

• Step 0: $n^2/e^n \to 0$ (exponential beats polynomial). Continue.
• Step 1: not geometric, $p$-series, or telescoping.
• Step 2: no alternating signs.
• Step 3: exponential present → ratio test. $$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{e^{n+1}}\cdot\frac{e^n}{n^2} = \frac{1}{e}\left(1 + \frac1n\right)^2 \to \frac1e < 1.$$ Converges.

Worked Example 22.10.B. Test $\displaystyle\sum_{n=2}^\infty \frac{1}{n \ln n}$.

• Step 0: terms $\to 0$. Continue.
• Step 1: not a named series.
• Step 2: not alternating.
• Step 3: no factorial/exponential (and the ratio $\to 1$, so the ratio test would be inconclusive anyway).
• Step 4: not an $n$-th power.
• Step 5: not a polynomial ratio — the $\ln n$ disqualifies limit comparison with a clean $p$-series.
• Step 6: decreasing with an easy integrand → integral test. As in §22.2, $\int_2^\infty \frac{dx}{x\ln x} = \infty$, so it diverges.

Worked Example 22.10.C. Test $\displaystyle\sum_{n=1}^\infty \frac{(-1)^n\, n}{n^2 + 1}$, and classify it.

• Step 0: $b_n = n/(n^2+1) \to 0$. Continue.
• Step 2: alternating → alternating series test. Is $b_n$ decreasing? Yes for $n \ge 1$ (the function $x/(x^2+1)$ decreases once $x > 1$, and you can confirm $b_{n+1}converges.
• Absolute or conditional? Test $\sum \frac{n}{n^2+1}$ by limit comparison with $1/n$: the ratio $\to 1$, and $\sum 1/n$ diverges, so $\sum |a_n|$ diverges. Therefore the series is conditionally convergent — a clean tour through Steps 0, 2, and 5 in one problem.

22.11 Summary and the Road to Taylor Series

You now hold the complete convergence toolkit. Here it is, organized by the signal that should trigger each test:

Three facts to carry forward above all others:

• The $p$-series rule — $\sum 1/n^p$ converges iff $p > 1$ — is your universal yardstick for limit comparison.
• The geometric series — $\sum r^n$ converges iff $|r| < 1$ — is the model behind both the ratio and root tests.
• Absolute convergence is order-independent; conditional convergence is not (Riemann). When in doubt, test $\sum |a_n|$ first.

Add to Your Modeling Portfolio. Add a convergence analysis to your model — identify an infinite sum it relies on and prove it is finite, then estimate how many terms you need. Biology: in an SIR-type or branching-process model, the expected total number of secondary infections is a series $\sum R_0^{\,n}$; use the geometric/ratio criterion to find the threshold $R_0 < 1$ that guarantees an epidemic dies out, and bound the total outbreak size. Economics: the present value of a perpetual cash flow is $\sum c_n/(1+r)^n$; show it converges for discount rate $r > 0$ and compute the finite present value via the geometric criterion. Physics: the energy of a vibrating string or the field of a charge lattice is an infinite sum; use comparison or the integral test to confirm the total energy is finite and bound the truncation error of a numerical mode sum. Data Science: many algorithms (e.g. gradient descent with a decaying step size $\eta_n$) converge iff $\sum \eta_n^2 < \infty$ while $\sum \eta_n = \infty$; use the $p$-series rule to choose a schedule like $\eta_n = 1/n^{0.7}$ that satisfies both conditions.

Looking ahead. Every test in this chapter asked a yes/no question about a fixed series of numbers. Chapter 23 changes the game: it studies power series $\sum c_n x^n$, whose terms depend on a variable $x$. The startling result is that the ratio and root tests you just learned, applied to a power series, carve out an interval of $x$-values — the radius of convergence — on which the series defines a genuine function. Inside that radius, you can differentiate and integrate term by term, and functions like $e^x$, $\sin x$, and $\ln(1+x)$ unfold into infinite polynomials. The alternating-series error bound you mastered in §22.7 becomes the tool that certifies how well a finite Taylor polynomial approximates the real thing — the very mechanism that lets a calculator compute $\sin(1)$ or that finishes the area-under-the-normal-curve story from Chapter 13. The tactical tests of this chapter are the foundation; Taylor series are the payoff.

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