Ratio and root tests

Hand a mathematician a series like $\sum 2^n/n!$ and ask "does it converge?" and the fastest honest answer rarely comes from inventing a clever comparison series — it comes from asking a single question: how does each term compare to the one before it, or to a fixed geometric shrinkage rate? The ratio test and the root test turn that question into a mechanical limit computation. Both tests secretly measure the same thing — how close a series is to a geometric series in disguise — which is why they agree whenever either gives a definite answer. This article builds both tests from that geometric-comparison idea, works a factorial series and an $n$ -th-power series side by side, and flags the one case both tests refuse to settle.

The core insight

A series converges absolutely if its terms eventually shrink at least as fast as a fixed geometric ratio less than 1, and the ratio and root tests are two different, equally valid ways of measuring that shrinkage rate.

The core idea

Ratio test. For $\sum a_n$ with eventually nonzero terms, compute the limit of consecutive term ratios:

L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|.

where: $a_n$ is the $n$ -th term of the series and $L$ is the limiting ratio (a nonnegative real number or $\infty$ ). If $L<1$ the series converges absolutely; if $L>1$ (including $L=\infty$ ) it diverges; if $L=1$ the test is inconclusive.

Root test. For $\sum a_n$ , compute the limit of the $n$ -th root of $|a_n|$ :

L = \lim_{n\to\infty} |a_n|^{1/n}.

where: the same three outcomes apply — $L<1$ converges absolutely, $L>1$ diverges, $L=1$ is inconclusive. The root test technically uses $\limsup$ so it is always defined even when the plain limit is not, but at this course's level the ordinary limit is what you compute.

Choosing between them. The ratio test wins whenever $a_n$ involves factorials or a product of $n$ linear/exponential factors, because $a_{n+1}/a_n$ triggers massive cancellation — $(n+1)!/n! = n+1$ collapses instantly. The root test wins whenever $a_n$ is itself raised to the $n$ -th power, $a_n = (f(n))^n$ , because the $n$ -th root simply undoes that power. Both tests are also the standard machinery for a power series' radius of convergence later in the chapter, via $1/R=\lim|a_{n+1}/a_n|$ or $1/R=\limsup|a_n|^{1/n}$ .

Figure

A number line showing consecutive terms of a shrinking series bounded above by a geometric sequence with ratio r less than one, illustrating why L less than one forces convergence

Why the tests work. If $|a_{n+1}/a_n|\to L<1$ , pick any $r$ with $L<r<1$ ; eventually $|a_{n+1}|\le r|a_n|$ , so beyond that point the terms are dominated by a geometric sequence with ratio $r$ , and $\sum a_n$ converges by direct comparison with $\sum r^n$ . The root test gives the same conclusion by a parallel route: $|a_n|^{1/n}\to L<1$ means eventually $|a_n|\le r^n$ , again a geometric bound. Both tests are, underneath, the same comparison-with-geometric-series argument stated two different ways.

Relative strength. Whenever $\lim|a_{n+1}/a_n|$ exists, it equals $\lim|a_n|^{1/n}$ , so the two tests never contradict each other on a definite answer. The root test is technically more powerful — it can decide some series where the ratio limit does not exist at all — but the ratio test is usually much faster to compute by hand, so it is the first tool reached for.

Derivation AWhy L < 1 forces absolute convergence under the ratio test

The goal is to show, starting only from the definition of a limit and the comparison test for series (both already available from earlier in this chapter), that $|a_{n+1}/a_n|\to L$ with $L<1$ is enough to guarantee $\sum |a_n|$ converges — the logical engine that makes the ratio test's first bullet point true rather than merely plausible.

Step 1. Fix a ratio strictly between $L$ and 1. Since $L<1$ , choose any real number $r$ with $L<r<1$ ; such an $r$ exists because there is room on the number line between $L$ and $1$ . Let $\varepsilon = r-L>0$ .

Step 2. Use the definition of the limit. Because $|a_{n+1}/a_n|\to L$ , there is an index $N$ such that for all $n\ge N$ , $\left||a_{n+1}/a_n|-L\right|<\varepsilon$ , which rearranges to $|a_{n+1}/a_n| < L+\varepsilon = r$ . In words: past index $N$ , every consecutive ratio is smaller than $r$ .

Step 3. Chain the inequality forward. For $n\ge N$ , $|a_{n+1}| < r\,|a_n|$ . Applying this repeatedly from $n=N$ : $|a_{N+1}|<r|a_N|$ , then $|a_{N+2}|<r|a_{N+1}|<r^2|a_N|$ , and in general $|a_{N+k}| < r^k |a_N|$ for every $k\ge 1$ .

Step 4. Compare with a geometric series. The tail $\sum_{k\ge1} |a_{N+k}|$ is termwise smaller than $|a_N|\sum_{k\ge1} r^k$ , which converges because $0<r<1$ (a geometric series with ratio between 0 and 1). By the comparison test, $\sum_{k\ge1}|a_{N+k}|$ converges, so the full series $\sum|a_n|$ (the finite head plus this convergent tail) converges. Hence $\sum a_n$ converges absolutely — exactly the claim.

Worked example

Test $\displaystyle\sum_{n=1}^{\infty} \frac{2^n}{n!}$ for convergence, and then decide the fate of the related series $\displaystyle\sum_{n=1}^{\infty} \left(\frac{3n+1}{4n-2}\right)^n$ .

Step 1 — set up the ratio for the factorial series. With $a_n = \dfrac{2^n}{n!}$ ,

\frac{a_{n+1}}{a_n} = \frac{2^{n+1}/(n+1)!}{2^n/n!} = \frac{2^{n+1}}{2^n}\cdot\frac{n!}{(n+1)!} = 2\cdot\frac{1}{n+1}.

The factorial almost entirely cancels — this is exactly the structure that makes the ratio test the right tool here rather than the root test.

Step 2 — take the limit.

L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{2}{n+1}=0.

Step 3 — conclude for the factorial series. Since $L=0<1$ , $\sum 2^n/n!$ converges absolutely. (This limit being $0$ also proves $2^n/n!\to0$ as $n\to\infty$ , which is why the terms of the Maclaurin series for $e^x$ eventually vanish for any fixed $x$ .)

Step 4 — switch tools for the $n$ -th-power series. With $a_n=\left(\dfrac{3n+1}{4n-2}\right)^n$ , a ratio-test attempt would require expanding $a_{n+1}/a_n$ into a messy quotient of two different $n$ -th powers. The root test instead undoes the power directly: $|a_n|^{1/n}=\dfrac{3n+1}{4n-2}$ .

Step 5 — take the limit and conclude. Dividing top and bottom by $n$ , $L=\lim_{n\to\infty}\dfrac{3+1/n}{4-2/n}=\dfrac{3}{4}$ . Since $L=3/4<1$ , this series also converges absolutely — reached in one line by picking the test that matches the term's structure.

Competitive stretch

What if the exponent were changed to $\left(\dfrac{3n+1}{2n-2}\right)^n$ ? Now $|a_n|^{1/n}\to \dfrac{3}{2}$ , so $L=3/2>1$ and the series diverges — the root test flips the verdict the instant the growth rate crosses $1$ , which is exactly the kind of boundary-sensitivity a JEE-style follow-up question likes to probe.

Common traps

Treating $L=1$ as an answer. $L=1$ means the test is silent, not that the series converges or diverges; $\sum 1/n$ and $\sum 1/n^2$ both give $L=1$ under the ratio test yet have opposite fates, so a different tool (comparison, integral test) is required.
Dropping the absolute value. Both tests are defined with $|a_{n+1}/a_n|$ and $|a_n|^{1/n}$ ; omitting it on series with negative or oscillating terms can produce a wrong or undefined limit.
Algebraic slips with factorials and powers. $(n+1)!/n!=n+1$ and $2^{n+1}/2^n=2$ are the two most common places to lose a factor; write the ratio out in full before cancelling anything.
Applying the ratio test where $a_n=0$ infinitely often. The ratio $a_{n+1}/a_n$ is undefined whenever $a_n=0$ , so such degenerate series need direct inspection before any test is attempted.

Ask the tutor: Not sure which test fits a given series, or want to see the root test resolve a case where the ratio test's limit does not even exist? Ask the AI tutor to walk through the choice on a few more examples.

Why do the ratio test and root test always agree when both give a definite answer?

Because whenever $\lim|a_{n+1}/a_n|$ exists, a standard limit fact guarantees $\lim|a_n|^{1/n}$ exists too and equals the same value. So on any series where the ratio limit is computable, you get the identical $L$ either way, and hence the identical verdict. They can only differ in edge cases where the ratio limit fails to exist but the root limit still does — which is why the root test is called the technically stronger of the two.

If the ratio test gives L equal to 1, does trying the root test help?

No, not directly — whenever the ratio limit exists and equals 1, the root limit also equals 1, so you get the same inconclusive result from both. $L=1$ under either test simply means neither geometric-comparison argument applies, and you must fall back on a different tool such as the comparison test, limit comparison test, or integral test, chosen based on the series' specific structure.

Do I need to prove the ratio test from scratch on an exam, or just apply it?

Almost all exam and homework problems only require applying the test: compute $L$ , state which case it falls into, and conclude. The proof (comparing the tail against a geometric series once the ratio drops below some $r<1$ ) is worth understanding once because it explains why the three cases are what they are, and it is exactly the argument you would reuse if asked to justify the test rather than just use it.

Ratio and root tests

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads