Root Test: Statement, Examples & When To Use It

What Is The Root Test?

The Root Test is a convergence test for an infinite series $\sum a_n$ that works by taking the $n$th root of the size of each term. You compute a single limit,

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

and read the answer straight off the value of $L$:

• Converges: if $L < 1$, the series converges absolutely (so it converges).

• Diverges: if $L > 1$, or $L = \infty$, the series diverges.

• Inconclusive: if $L = 1$, the test says nothing, and you must try a different test.

The absolute value bars matter. The Root Test measures $|a_n|$, so a positive result proves absolute convergence, which is the strongest kind and holds whether the terms are positive, negative, or mixed. The test is also called the Cauchy root test, after the mathematician who first wrote it down. For where it sits among the other tools, see convergence and divergence of series.

How Do You Apply The Root Test Step By Step?

The whole method is three moves. The only real skill is the limit in the middle.

1. Form $|a_n|$, the absolute value of the general term.

2. Raise it to the power $1/n$ and simplify. When $a_n$ is already an $n$th power, the exponent $1/n$ cancels the outer power cleanly, which is exactly why this test loves $n$th-power terms.

3. Take the limit as $n \to \infty$ to get $L$, then apply the three cases above.

The reason step 2 is so tidy for $n$th powers is worth seeing once. If $a_n = (b_n)^n$, then

$$|a_n|^{1/n} = \big|(b_n)^n\big|^{1/n} = |b_n|,$$

so the whole test collapses to $\lim_{n \to \infty} |b_n|$. The messy outer power vanishes and you are left with the base. That single cancellation is what makes the Root Test the natural first choice when you spot a term raised to the $n$th power.

What Are Worked Examples Of The Root Test?

Each example is fully stepped, and each ends with the case it lands in.

Example 1: A term raised to the $n$th power (converges).

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

The term is an $n$th power, so the $1/n$ exponent cancels it:

$$|a_n|^{1/n} = \left[\left(\frac{n}{2n+1}\right)^{n}\right]^{1/n} = \frac{n}{2n+1}.$$

Now take the limit by dividing numerator and denominator by $n$:

$$L = \lim_{n \to \infty} \frac{n}{2n+1} = \lim_{n \to \infty} \frac{1}{2 + \tfrac{1}{n}} = \frac{1}{2}.$$

Since $L = \tfrac{1}{2} < 1$, the series converges.

Final answer: $L = \tfrac{1}{2} < 1$, so the series converges absolutely.

Example 2: A power over a power (converges).

Test $\displaystyle\sum_{n=1}^{\infty} \frac{2^{n}}{n^{n}}$.

Both the numerator and the denominator are $n$th powers, so the $n$th root reaches the whole term at once:

$$|a_n|^{1/n} = \left(\frac{2^{n}}{n^{n}}\right)^{1/n} = \frac{2}{n}.$$

Then

$$L = \lim_{n \to \infty} \frac{2}{n} = 0.$$

Since $L = 0 < 1$, the series converges. Notice the root had to act on both the $2^n$ and the $n^n$; taking it on only one part is the classic slip covered in the mistakes section.

Final answer: $L = 0 < 1$, so the series converges absolutely.

Example 3: A term raised to the $n$th power (diverges).

Test $\displaystyle\sum_{n=1}^{\infty} \left(\frac{3n+1}{2n+5}\right)^{n}$.

Cancel the outer power with the $n$th root:

$$|a_n|^{1/n} = \frac{3n+1}{2n+5}.$$

Divide through by $n$ and take the limit:

$$L = \lim_{n \to \infty} \frac{3 + \tfrac{1}{n}}{2 + \tfrac{5}{n}} = \frac{3}{2}.$$

Since $L = \tfrac{3}{2} > 1$, the terms do not shrink toward zero fast enough, and the series diverges.

Final answer: $L = \tfrac{3}{2} > 1$, so the series diverges.

Example 4: When the test cannot decide (inconclusive).

Test the $p$-series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{p}}$ for any fixed $p$.

Here $|a_n|^{1/n} = \left(n^{-p}\right)^{1/n} = \left(n^{1/n}\right)^{-p}$. The key fact is that $\lim_{n \to \infty} n^{1/n} = 1$, because

$$\ln\left(n^{1/n}\right) = \frac{\ln n}{n} \longrightarrow 0 \quad\Rightarrow\quad n^{1/n} \longrightarrow e^{0} = 1.$$

Therefore

$$L = \lim_{n \to \infty} \left(n^{1/n}\right)^{-p} = 1^{-p} = 1.$$

The Root Test returns $L = 1$ for every $p$, so it can never tell you that $\sum \tfrac{1}{n^2}$ converges while $\sum \tfrac{1}{n}$ diverges, even though both facts are true. This is the honest limit of the test: on $p$-series it is silent, and you switch to the integral test or a comparison test instead.

Final answer: $L = 1$, so the Root Test is inconclusive here.

Table: The four worked examples at a glance.

| Series | $|a_n|^{1/n}$ | $L$ | Verdict | |---|---|---|---| | $\sum \left(\dfrac{n}{2n+1}\right)^n$ | $\dfrac{n}{2n+1}$ | $\tfrac{1}{2}$ | Converges | | $\sum \dfrac{2^n}{n^n}$ | $\dfrac{2}{n}$ | $0$ | Converges | | $\sum \left(\dfrac{3n+1}{2n+5}\right)^n$ | $\dfrac{3n+1}{2n+5}$ | $\tfrac{3}{2}$ | Diverges | | $\sum \dfrac{1}{n^p}$ | $\left(n^{1/n}\right)^{-p}$ | $1$ | Inconclusive |

When Should You Use The Root Test Instead Of The Ratio Test?

The Root Test and the ratio test answer the same yes-or-no question and share the same three cutoffs at $1$. Choosing between them is the question readers ask most, so here is the honest comparison.

• Reach for the Root Test when the general term is raised to the $n$th power, such as $\left(\tfrac{n}{2n+1}\right)^n$ or $\tfrac{2^n}{n^n}$. The $1/n$ exponent cancels the power and the limit falls out in one line.

• Reach for the ratio test when the term contains a factorial or a product that telescopes when you form $\tfrac{a_{n+1}}{a_n}$. Factorials cancel beautifully in a ratio and awkwardly under a root.

There is also a genuine difference in strength. The Root Test is more powerful than the ratio test: whenever the ratio test gives a verdict, the Root Test gives the same verdict, and there are series the Root Test settles while the ratio test returns $1$ and gives up. The reason is a standard inequality between the two limits,

$$\lim_{n \to \infty} \sqrt[n]{|a_n|} \le \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$$

(read with $\liminf$ and $\limsup$ when the plain limits do not exist). So the Root Test never loses to the ratio test on strength. In day-to-day work the ratio test is often the easier arithmetic, which is why both stay in the toolkit.

How Does The Root Test Find The Radius Of Convergence?

The Root Test does more than label a fixed series. Applied to a power series $\sum c_n x^n$, it hands you the exact interval on which that series converges.

Apply the test to the term $a_n = c_n x^n$:

$$L = \lim_{n \to \infty} |c_n x^n|^{1/n} = |x| \cdot \lim_{n \to \infty} |c_n|^{1/n}.$$

Write $\ell = \lim_{n \to \infty} |c_n|^{1/n}$. The series converges when $L < 1$, that is when $|x| , \ell < 1$, or $|x| < \tfrac{1}{\ell}$. That threshold is the radius of convergence:

$$R = \frac{1}{\displaystyle\lim_{n \to \infty} |c_n|^{1/n}}.$$

This is the Cauchy-Hadamard theorem, and it is why the Root Test is the theoretical backbone of power series, not just one test among many. The same $n$th-root limit that decides a single sum also draws the boundary of every Taylor and power series.

Why Does The Root Test Work?

The whole test is a comparison with a geometric series, and the geometric picture is the intuition behind the algebra.

• The convergent case is a shrinking staircase. If $L < 1$, pick any number $r$ between $L$ and $1$. For large $n$ the root $|a_n|^{1/n}$ sits below $r$, which means $|a_n| < r^n$. Each term is eventually smaller than a step of a geometric series with ratio $r < 1$, and a geometric series with ratio below $1$ has a finite total. By the comparison test, the tail of $\sum |a_n|$ is trapped under a finite geometric sum, so it converges.

• The divergent case fails the smell test. If $L > 1$, then $|a_n|^{1/n} > 1$ for large $n$, so $|a_n| > 1$ over and over. The terms do not shrink toward zero, and a series whose terms do not approach zero cannot converge, by the divergence test.

• The boundary is genuinely undecided. If $L = 1$, the terms are borderline: they might shrink just fast enough to converge, like $\tfrac{1}{n^2}$, or just too slowly, like $\tfrac{1}{n}$. Both give $L = 1$, so the root alone cannot separate them.

Read as one idea, the Root Test asks whether $a_n$ eventually behaves like $r^n$ for some fixed ratio $r$. If that hidden ratio is below $1$, the sum is finite; if it is above $1$, the sum runs away; and exactly at $1$, the test is looking at the one case a geometric comparison cannot resolve.

Who Discovered The Root Test?

The test grew out of the first serious attempt to put infinite series on solid ground, rather than treating them as informal sums.

One later refinement is worth naming. Cauchy's original version assumed the limit $\lim_{n \to \infty} |a_n|^{1/n}$ exists. Analysts after him sharpened it to use the limit superior ($\limsup$), the largest value the roots keep returning to, so that the test still applies when the ordinary limit wobbles and never settles. The modern statement you meet in a rigorous course is this sharper $\limsup$ form.

Where Is The Root Test Used In The Real World?

The Root Test rarely appears by name outside a classroom, but the convergence question it answers sits under a surprising amount of applied mathematics.

• Function approximation: engineers and physicists replace hard functions with power series, and the Cauchy-Hadamard radius, set by the same $n$th-root limit, tells them the range of inputs where that approximation is trustworthy.

• Signal processing: the $z$-transform of a digital signal is a power series in disguise, and its region of convergence, found by a root-style limit, decides whether a filter is stable.

• Control and stability: deciding whether repeated feedback in a system dies out or blows up is a convergence question of exactly the form the Root Test settles.

• Numerical methods: an iterative algorithm is worth running only if its error series converges, and comparing successive errors to a fixed ratio is the same shrinking-geometric idea the Root Test formalises.

• Finance: the present value of a stream of shrinking future payments is a geometric-style series, and it has a finite, usable total precisely when the effective ratio stays below $1$.

One limit about $n$th roots quietly guards the line between models that give finite, usable answers and models that run away to infinity.

What Are The Most Common Mistakes With The Root Test?

These three errors account for most lost marks on the Root Test, and each matches a question real students ask on r/calculus, r/learnmath, and course resource centres.

Taking the $n$th root of only part of the term.

Where it slips in:

On $\sum \tfrac{2^n}{n^n}$ a student writes $|a_n|^{1/n} = \tfrac{2}{n^n}$ or $\tfrac{2^n}{n}$, rooting the numerator or the denominator but not both.

Don't do this:

Do not apply the $1/n$ power to a fragment. The root acts on the whole term $|a_n|$ at once.

The correct way:

Root everything together: $\left(\tfrac{2^n}{n^n}\right)^{1/n} = \tfrac{2}{n}$, which then tends to $0$. Both powers get cancelled, not just one.

Reading a verdict out of $L = 1$.

Where it slips in:

A student computes $L = 1$ for a $p$-series and concludes it converges (or diverges), treating the boundary value as an answer.

Don't do this:

Do not decide anything when $L = 1$. The Root Test is genuinely silent there, for both convergent and divergent series.

The correct way:

Switch tests. When $L = 1$, move to the integral test, a comparison test, or the limit comparison test, which can separate the borderline cases the root cannot.

Dropping the absolute value.

Where it slips in:

On a series with negative or alternating terms, a student roots $a_n$ directly and hits the $n$th root of a negative number, or silently ignores the sign.

Don't do this:

Do not root a signed term. The test is defined on $|a_n|$, not $a_n$.

The correct way:

Always take the size first: compute $\lim_{n \to \infty} |a_n|^{1/n}$. A result below $1$ proves absolute convergence, which is exactly why the bars are part of the definition.

Practice Problems On The Root Test

Work each one, then check against the answer. Answers are verified.

1. Test $\displaystyle\sum_{n=1}^{\infty} \left(\frac{n+4}{5n+1}\right)^{n}$.
   (Answer: $L = \lim \tfrac{n+4}{5n+1} = \tfrac{1}{5} < 1$, converges.)

2. Test $\displaystyle\sum_{n=1}^{\infty} \frac{3^{n}}{n^{n}}$.
   (Answer: $|a_n|^{1/n} = \tfrac{3}{n} \to 0 < 1$, converges.)

3. Test $\displaystyle\sum_{n=1}^{\infty} \left(\frac{4n-1}{3n+2}\right)^{n}$.
   (Answer: $L = \tfrac{4}{3} > 1$, diverges.)

4. Test $\displaystyle\sum_{n=1}^{\infty} \frac{n^{n}}{5^{n}}$.
   (Answer: $|a_n|^{1/n} = \tfrac{n}{5} \to \infty > 1$, diverges.)

5. Test $\displaystyle\sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)^{n^{2}}$.
   (Answer: $|a_n|^{1/n} = \left(1 - \tfrac{1}{n}\right)^{n} \to e^{-1} \approx 0.3679 < 1$, converges.)

6. Test $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(\ln n)^{n}}$ (from $n = 2$).
   (Answer: $|a_n|^{1/n} = \tfrac{1}{\ln n} \to 0 < 1$, converges.)

Where Should You Go Next After The Root Test?

The Root Test is one hinge in the wider study of series, and several natural doors open from here.

1. Ratio test. Learn the sibling test that handles factorials with ease, and see exactly where the two agree and where the root wins.

2. Radius of convergence. Turn the Root Test on a power series and read off the interval where it is valid, via Cauchy-Hadamard.

3. Power series. Build the object the Root Test was quietly designed to analyse, the foundation of Taylor approximations.

If your child is meeting convergence tests for the first time, a live Bhanzu trainer teaches the Root Test alongside its neighbours so the choice of test feels natural rather than memorised, in the Bhanzu math program.