The Root Test

Many infinite sums are built from terms raised to a growing power — a rate compounded over and over, or a signal fading by a fixed fraction at every step. To decide whether such a sum stays finite, the root test peels that power away in a single clean move.

Every convergence test is secretly asking the same question: does this series behave like a geometric series with ratio below 1? The ratio test answers it by comparing each term to the one before. The root test is more direct: it takes the nth root of the nth term and reads the term's effective ratio straight off its size. If |a_n| \approx r^n in the long run, then \sqrt[n]{|a_n|} \approx r — the root recovers r directly, no neighbouring term needed.

L = \lim_{n\to\infty} \sqrt[n]{\,|a_n|\,} = \lim_{n\to\infty} |a_n|^{1/n}.

L < 1 — the series converges absolutely: its terms are eventually beaten by a geometric r^n.
L > 1 (or L = \infty) — it diverges: the terms don't even tend to zero.
L = 1 — the test is silent. No verdict either way.

And here is the reason the root test earns its place next to the ratio test: it eats anything with an n in the exponent. A term like \big(\tfrac{2n}{3n+1}\big)^{n} is a nightmare to divide by its predecessor, but one nth root peels the exponent clean off. Whenever the whole term is "something to the power n", the root test is not just an option — it is the tool the term was shaped for.

Two little limits you will use every time

Taking nth roots sounds fearsome until you know the two facts that make almost every computation collapse. Both come from writing x^{1/n} = e^{(\ln x)/n}:

\sqrt[n]{c} \;=\; c^{1/n} \longrightarrow 1 \quad\text{for any constant } c > 0, \qquad\qquad \sqrt[n]{n} \;=\; n^{1/n} \longrightarrow 1.

The first holds because \tfrac{\ln c}{n} \to 0; the second because \tfrac{\ln n}{n} \to 0 — the logarithm grows far too slowly to keep up with n. Chaining them gives \sqrt[n]{n^k} = \big(n^{1/n}\big)^k \to 1 for any fixed power k, and then \sqrt[n]{5n^3 + 7} \to 1 too.

The moral: polynomial factors are invisible to the root test. Constants, powers of n, whole polynomials — the nth root flattens them all to 1. Only the exponential-scale part of a term survives the root, which is exactly why the test measures a term's effective geometric ratio and nothing else.

Worked examples

Example 1 — the tailor-made shape: \displaystyle\sum_{n=1}^{\infty} \left(\frac{2n}{3n+1}\right)^{n}. The whole term is an nth power, so the root strips it in one move:

|a_n|^{1/n} = \left(\left(\frac{2n}{3n+1}\right)^{n}\right)^{1/n} = \frac{2n}{3n+1} \;\longrightarrow\; \frac{2}{3}.

So L = \tfrac23 < 1: the series converges absolutely. Notice there was no algebra worth the name — the exponent n and the root 1/n cancel, leaving a limit any first-year student can read off. Try writing a_{n+1}/a_n for this series and you will appreciate the shortcut.

Example 2 — a base that runs away: \displaystyle\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{n}}. Again the term is an nth power, this time of a shrinking base:

|a_n|^{1/n} = \frac{1}{\ln n} \;\longrightarrow\; 0.

Here L = 0 < 1, the strongest possible verdict: eventually the terms are smaller than r^n for every r > 0, however tiny. The series converges absolutely — and ferociously fast. (Don't be spooked that \ln n grows slowly; it still grows, and that is all the root needs.)

Example 3 — where the ratio would be ugly but the root is instant: \displaystyle\sum_{n=1}^{\infty} \frac{1}{3^{n}}\left(\frac{n+1}{n}\right)^{n^{2}}. The ratio a_{n+1}/a_n pits \big(\tfrac{n+2}{n+1}\big)^{(n+1)^2} against \big(\tfrac{n+1}{n}\big)^{n^2} — a genuinely horrible expression. The root, by contrast, just divides the exponent n^2 by n:

|a_n|^{1/n} = \frac{1}{3}\left(\frac{n+1}{n}\right)^{n} = \frac{1}{3}\left(1 + \frac{1}{n}\right)^{n} \;\longrightarrow\; \frac{e}{3} \approx 0.906.

Since e < 3, we get L = e/3 < 1 and the series converges absolutely — with the famous limit (1+\tfrac1n)^n \to e making a guest appearance. A tight verdict, too: nudge the 3 down to e and the test would fall silent.

The division of labour is worth memorising as a reflex: nth powers \Rightarrow root test; factorials \Rightarrow ratio test. A factorial telescopes beautifully in a ratio (\tfrac{(n+1)!}{n!} = n+1) but resists an nth root; an nth power collapses under a root but makes a mess in a ratio. Pick the tool that matches the shape.

Why L < 1 forces convergence

As with the ratio test, the trick is to trap the tail under a convergent geometric series — only here a single root, not a chain of ratios, does the work. Suppose L = \lim_{n\to\infty} |a_n|^{1/n} < 1.

Step 1 — choose r between L and 1.

L < r < 1.

Step 2 — the roots eventually drop below r. Since |a_n|^{1/n} \to L < r, past some index N the roots never exceed r:

|a_n|^{1/n} \le r \qquad \text{for all } n \ge N.

Step 3 — raise both sides to the nth power. This is the move the root test is built for — the exponent comes right off:

|a_n| \le r^{\,n} \qquad \text{for all } n \ge N.

Step 4 — dominate the tail by a geometric series. Sum from N onward and compare term by term to a geometric series of ratio r < 1:

\sum_{n=N}^{\infty} |a_n| \;\le\; \sum_{n=N}^{\infty} r^{\,n} = \frac{r^{N}}{1 - r} < \infty.

Step 5 — conclude absolute convergence. The tail is finite, and the finitely many earlier terms add only a finite amount, so

\sum_{n=1}^{\infty} |a_n| < \infty.

If L > 1, then |a_n|^{1/n} > 1 infinitely often, so |a_n| > 1 infinitely often — the terms do not tend to zero, and the divergence test finishes the job.

For a series \sum a_n, let L = \displaystyle\lim_{n\to\infty} |a_n|^{1/n} (use \limsup if the plain limit does not exist — the \limsup always does). Then:

L < 1 \Rightarrow \sum a_n converges absolutely (tail dominated by \sum r^n, r \in (L,1)).
L > 1 or L = \infty \Rightarrow \sum a_n diverges (terms do not tend to zero).
L = 1 \Rightarrow inconclusive (e.g. every p-series gives L = 1).

L = 1 is silence, not "converges slowly". And it is silence on exactly the series students most want to settle: the harmonic series \sum \tfrac1n (diverges) and \sum \tfrac1{n^2} (converges) both give |a_n|^{1/n} \to 1, because n^{1/n} \to 1. Worse: this is the same stalemate zone as the ratio test, on the same examples — whenever the ratio limit exists the root limit equals it, so if the root test says 1, switching to the ratio test cannot rescue you. Reach for the comparison or integral test instead.
Factorials are the root test's blind spot. What is \sqrt[n]{n!}? Not obvious — you need Stirling's approximation (\sqrt[n]{n!} \sim n/e) to make progress, which is heavier machinery than the problem deserves. The ratio test dissolves the same factorial in one line via \tfrac{(n+1)!}{n!} = n+1. Shape-match your tool: factorials \to ratio, nth powers \to root.
In full rigor the test uses \limsup. The sequence |a_n|^{1/n} need not converge at all, but its \limsup always exists (allowing \infty), and the verdicts \limsup < 1 / \limsup > 1 stand unchanged. When the plain limit exists it equals the \limsup, so in everyday calculations you may write \lim with a clear conscience.

The two tests usually agree, and when both ratios and roots converge they give the same L — there is a theorem that

\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = L \;\Longrightarrow\; \lim_{n\to\infty} |a_n|^{1/n} = L,

but not conversely. The root test can succeed where the ratio test stalls. The classic witness has terms that alternate two recipes,

a_n = \begin{cases} 2^{-n}, & n \text{ even},\\[2pt] 3^{-n}, & n \text{ odd}. \end{cases}

The ratios a_{n+1}/a_n swing wildly between huge and tiny and never settle, so the ratio test has no limit to read. But the roots are tame — |a_n|^{1/n} is either \tfrac12 or \tfrac13, so \limsup = \tfrac12 < 1 and convergence is immediate. So in theory the root test strictly dominates: every series the ratio test settles, the root test settles too, and not vice versa. Yet in practice the ratio test gets far more use — ratios are simply easier to compute for the factorial-laden series that fill calculus courses. Strength in principle, convenience in practice: keep both in the toolbox.

The root test is not just one test among many — it is the engine under power series. Apply it to \sum c_n x^n: the nth root of |c_n x^n| is |x| \cdot |c_n|^{1/n}, so with \ell = \limsup |c_n|^{1/n} the series converges absolutely whenever |x| \ell < 1 and diverges whenever |x| \ell > 1. That is the Cauchy–Hadamard formula for the radius of convergence:

R = \frac{1}{\limsup_{n\to\infty} |c_n|^{1/n}}.

Because the \limsup always exists, every power series — however wild its coefficients — gets a clean disc of convergence, with no exceptions. The test itself is often called Cauchy's root test after Augustin-Louis Cauchy, who published it in his 1821 Cours d'analyse; the radius formula was stated by him too, forgotten, and rediscovered in 1888 by the 22-year-old Jacques Hadamard in his very first paper.

Watch the root find its limit

The whole test is one curve against one dashed line. Pick a series and watch |a_n|^{1/n} approach its limit L against the verdict line at 1. For \big(\tfrac{n}{2n+1}\big)^{n} the root is simply \tfrac{n}{2n+1} \to \tfrac12 < 1 — it settles comfortably below the line and the series converges. For 1/n^2 the root n^{-2/n} creeps up towards the line and lands exactly on it — the inconclusive L = 1, even though this series happens to converge. For (3/2)^n the root sits flat at 3/2 > 1 from the very first term — divergence, no waiting required.

Notice how slowly the middle curve makes up its mind — that crawl towards 1 is the visual signature of a series living at polynomial scale, where the root test has nothing to say. The decisive cases are the ones that settle cleanly away from the line.