Comparison Tests

Take a medicine at the same time each day and each dose leaves a little behind; does the amount in your body build to a safe steady level or climb without limit? Questions like this are really asking whether an infinite sum settles or runs away — and the quickest way to decide is to hold your messy series up against a simpler one whose fate you already know.

Judge a series by the company it keeps. If a series with positive terms sits underneath a series you already know converges, it has nowhere to escape to — it converges too. If it sits above a series you know diverges, it gets dragged off to infinity with it. That single idea — squeeze your unknown series between a known one and infinity — powers the two most-used convergence tests in the whole subject.

0 \le a_n \le b_n: \qquad \sum b_n \text{ converges} \Rightarrow \sum a_n \text{ converges}, \qquad \sum a_n \text{ diverges} \Rightarrow \sum b_n \text{ diverges}.

Notice what this means in practice: the whole game is picking the right benchmark. The test itself is almost trivial; the skill is choosing what to compare against. Two families act as the rulers of the trade, because their fate is known completely:

\text{geometric: } \sum r^n \text{ converges} \iff |r| < 1, \qquad \text{p-series: } \sum \frac{1}{n^p} \text{ converges} \iff p > 1.

The geometric series rules everything with exponential behaviour; the p-series (settled by the integral test) rules everything with power-of-n behaviour. Nearly every comparison you will ever run measures a messy series against one of these two rulers.

The Direct Comparison Test, worked line by line

Suppose 0 \le a_n \le b_n for all n, and that the larger series \sum b_n converges. We show \sum a_n converges — and the proof is short enough to carry in your pocket.

Step 1 — partial sums are increasing. Every term a_n \ge 0, so adding more terms only grows the partial sum s_n = \sum_{k=1}^{n} a_k:

s_1 \le s_2 \le s_3 \le \cdots

Step 2 — bound them above, term by term. Since a_k \le b_k for each k, the sum of the small terms is at most the sum of the big ones, which is at most the whole convergent total B = \sum_{k=1}^{\infty} b_k:

s_n = \sum_{k=1}^{n} a_k \;\le\; \sum_{k=1}^{n} b_k \;\le\; \sum_{k=1}^{\infty} b_k = B.

Step 3 — invoke monotone convergence. The sequence (s_n) is increasing (Step 1) and bounded above by B (Step 2). An increasing sequence with a ceiling must converge to a limit no larger than that ceiling:

\sum_{n=1}^{\infty} a_n = \lim_{n\to\infty} s_n \le B < \infty.

The divergence half is the contrapositive: if \sum a_n = \infty, the bigger \sum b_n cannot be finite either.

Suppose 0 \le a_n \le b_n for all n (or all n \ge N). Then:

\sum b_n converges \;\Rightarrow\; \sum a_n converges (and \sum a_n \le \sum b_n).
\sum a_n diverges \;\Rightarrow\; \sum b_n diverges.
The other two directions — smaller than a divergent series, or bigger than a convergent one — conclude nothing.

Worked example: below a convergent ruler

Test \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 3}. Step 1 — x-ray the term. For large n the +3 is a rounding error next to n^2, so the term behaves like 1/n^2. That nominates the ruler. Step 2 — check the inequality points the useful way. Adding 3 to the denominator makes the fraction smaller:

0 \le \frac{1}{n^2 + 3} \le \frac{1}{n^2} \qquad \text{for every } n \ge 1.

Step 3 — conclude. The ruler \sum 1/n^2 converges (p-series, p = 2 > 1), and our series sits underneath it, so by direct comparison it converges. As a bonus the proof hands us a bound: \sum 1/(n^2+3) \le \sum 1/n^2 = \pi^2/6 \approx 1.645. Three lines, done — when the inequality cooperates.

When the inequality points the wrong way

Now try \displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - 1}. Same x-ray, same instinct: it behaves like 1/n^2, so it surely converges. But watch the inequality. Subtracting 1 from the denominator makes the fraction bigger:

\frac{1}{n^2 - 1} \;\ge\; \frac{1}{n^2}.

Our series is bigger than a convergent series — and that is one of the two useless directions. A thing larger than something finite can still be finite or infinite; the comparison proves nothing at all. The test hasn't failed; we've simply pointed it the wrong way, like trying to prove someone is short by showing they're taller than a short person.

You can rescue direct comparison by hand-crafting a cleverer inequality — for n \ge 2 we have n^2 - 1 \ge n^2/2, hence 1/(n^2-1) \le 2/n^2, and twice a convergent series still converges. But notice what that took: a small flash of ingenuity, a constant conjured from nowhere, an inequality to verify. Multiply that fuss across every messy series you'll ever meet and you want a tool that skips it. That tool is the limit comparison test: it only asks whether the terms are eventually proportional to the ruler's, and never asks which one is on top.

The Limit Comparison Test, worked line by line

Let a_n, b_n > 0 and suppose the ratio of terms settles down to a finite, positive limit:

\lim_{n\to\infty} \frac{a_n}{b_n} = c, \qquad 0 < c < \infty.

Step 1 — turn the limit into a two-sided bound. The ratio settling on a positive c means that, past some index N, the ratio is trapped in a band around c — say between \tfrac{c}{2} and 2c:

\frac{c}{2} \;<\; \frac{a_n}{b_n} \;<\; 2c \qquad \text{for all } n \ge N.

Step 2 — multiply through by b_n > 0. This sandwiches the terms a_n between two constant multiples of b_n:

\frac{c}{2}\, b_n \;<\; a_n \;<\; 2c\, b_n \qquad \text{for all } n \ge N.

Step 3 — apply direct comparison both ways. A nonzero constant multiple does not change whether a series converges, so:

\sum b_n \text{ converges} \;\overset{a_n < 2c\,b_n}{\Longrightarrow}\; \sum a_n \text{ converges}, \qquad \sum b_n \text{ diverges} \;\overset{a_n > \frac{c}{2}\,b_n}{\Longrightarrow}\; \sum a_n \text{ diverges}.

Step 4 — conclude the shared fate. Both implications run, so the two series stand or fall together:

\sum a_n \text{ converges} \;\iff\; \sum b_n \text{ converges}.

Let a_n, b_n > 0 and \displaystyle\lim_{n\to\infty} \frac{a_n}{b_n} = c. Then:

If 0 < c < \infty, then \sum a_n and \sum b_n both converge or both diverge.
If c = 0, only one direction survives: \sum b_n converges \Rightarrow \sum a_n converges.
If c = \infty, only the other: \sum b_n diverges \Rightarrow \sum a_n diverges.

Worked example: a messy fraction of polynomials

Test \displaystyle\sum_{n=1}^{\infty} \frac{2n + 1}{n^3 + n}. Step 1 — x-ray. Keep only the dominant terms: top \approx 2n, bottom \approx n^3, so the term behaves like 2n/n^3 = 2/n^2. Ruler: b_n = 1/n^2. Step 2 — compute the ratio.

\frac{a_n}{b_n} = \frac{(2n+1)/(n^3+n)}{1/n^2} = \frac{(2n+1)\,n^2}{n^3+n} = \frac{2n^3 + n^2}{n^3 + n} = \frac{2 + \tfrac{1}{n}}{1 + \tfrac{1}{n^2}} \;\xrightarrow[n\to\infty]{}\; \frac{2 + 0}{1 + 0} = 2.

Step 3 — conclude. The limit c = 2 is finite and strictly positive, and the ruler \sum 1/n^2 converges, so the original series converges. No inequality was checked anywhere — the limit did all the work. (And c = 2 is no accident: it is exactly the leading coefficient the x-ray predicted.)

Worked example: the rescue

Back to the series that embarrassed direct comparison, \displaystyle\sum_{n=2}^{\infty} \frac{1}{n^2 - 1} with ruler b_n = 1/n^2:

\frac{a_n}{b_n} = \frac{1/(n^2-1)}{1/n^2} = \frac{n^2}{n^2 - 1} = \frac{1}{1 - \tfrac{1}{n^2}} \;\xrightarrow[n\to\infty]{}\; 1.

c = 1 \in (0, \infty) and \sum 1/n^2 converges, so \sum 1/(n^2-1) converges. The key lesson of this page in one sentence: when your series is a hair's breadth above its natural ruler, direct comparison points the wrong way — and limit comparison doesn't care.

How to pick the ruler

Every comparison argument starts the same way: stare at a_n and ask "what does this really look like as n \to \infty?" The reliable recipe:

Keep only the dominant term on top and bottom — the fastest-growing one. Constants, lower powers, lonely +7s: all noise.
Simplify what's left. Powers of n collapse to a p-series 1/n^p; exponentials collapse to a geometric series r^n. (Exponentials dominate every power, so 2^n + n^{100} x-rays to 2^n.)
Pick the test. If the inequality against the ruler is obvious and points the useful way, direct comparison is a one-liner. If not — don't wrestle it — go straight to limit comparison and compute c.

Because the leftovers vanish in the ratio. Take a_n = \dfrac{3n + 5}{n^3 - 2n + 4}: the x-ray says 3n/n^3 = 3/n^2, so compare with b_n = 1/n^2:

\frac{a_n}{b_n} = \frac{(3n + 5)\,n^2}{n^3 - 2n + 4} = \frac{3 + \tfrac{5}{n}}{1 - \tfrac{2}{n^2} + \tfrac{4}{n^3}} \;\xrightarrow[n\to\infty]{}\; 3.

Every discarded term reappears divided by a power of n and dies, leaving exactly the ratio of leading coefficients — a finite, positive c, which is precisely what limit comparison demands. The general rule of thumb: a rational term with degree e on top and degree d on the bottom behaves like 1/n^{\,d-e}, so it converges iff d - e > 1. This is why limit comparison is the workhorse for messy fractions of polynomials: the heuristic that picks the ruler also guarantees the test will apply.

Ask a working analyst whether \sum \frac{n^2 + 7n}{n^4 + \sqrt{n} + 9} converges and you'll get an answer before you finish writing it. They aren't computing — they are degree-counting: top grows like n^2, bottom like n^4, difference 4 - 2 = 2 > 1 — converges. It's the same x-ray habit you've just learned, compiled into reflex. Try reading these at a glance:

\sum \dfrac{n}{n^3 + 5} → behaves like 1/n^2 → converges.
\sum \dfrac{n^2 + 1}{n^3 + n} → behaves like 1/n → diverges (a harmonic series in a trench coat).
\sum \dfrac{1}{2^n + n^{10}} → the exponential owns the denominator → behaves like (1/2)^n → converges.
\sum \dfrac{1}{\sqrt{n^3 + n}} → behaves like 1/n^{3/2} → converges (p = 3/2 > 1 — fractional degrees count too).

The eyeball verdict is not a proof — the limit comparison computation is the proof — but in practice the x-ray is right so often that analysts treat degree-counting as a superpower: see the dominant term, know the fate, then write the two-line limit to make it official.

Three traps catch nearly every newcomer to comparison tests:

The inequality must point the right way — two directions are useless. Being smaller than a divergent series proves nothing: 1/n^2 \le 1/n and \sum 1/n diverges, yet \sum 1/n^2 converges happily. Being bigger than a convergent series proves nothing either: 1/n \ge 1/n^2 and \sum 1/n^2 converges, yet \sum 1/n diverges anyway. Only two directions carry information: below a convergent ceiling, or above a divergent floor.
Limit comparison needs c strictly between 0 and \infty. If the ratio drifts to 0 or blows up to \infty, the two series are on different scales and only a one-way conclusion survives (see the theorem's edge cases). Getting c = 0 usually means you picked a ruler that's too big — x-ray again.
The terms must be positive (at least eventually). Both proofs lean on partial sums that only ever increase; a series with mixed signs can sneak its partial sums back down and the whole trapping argument collapses. For \sum (-1)^n a_n you need different tools entirely — comparison tests there apply only to the series of absolute values.

Watch one series bound another

Here is the direct comparison test made visible. The faint dashed curve is the convergent yardstick b_n = 1/n^2; the bold curve is a_n = 1/(n^2 + k), which sits at or below it for every k \ge 0. Slide k down to 0 and the two curves fuse; crank it up to 20 and the bold curve is pushed further below the ceiling — but it can never poke above it.

The second chart is where the theorem actually lives: the partial sums. The dashed running total climbs towards \pi^2/6 \approx 1.645 and flattens out; the bold running total climbs too (Step 1 of the proof: it only ever increases) but stays pinned beneath the dashed one (Step 2: term-by-term smaller means sum-by-sum smaller). An increasing curve under a finite ceiling has no choice but to level off — that levelling-off is convergence, and it's Step 3 happening before your eyes.