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The Root Test

Calculus 2 · Chapter 7: Convergence Tests for SeriesSection 7.6 · lesson 37 of 549 min · free

← The Ratio TestNext lesson: Strategy for Testing Series →

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In this lesson

When the whole term is raised to the nth power.

  • Compute L = \lim|a_{n}|^{1/n} and read off the verdict
  • Use the two standard limits that do most of the work
  • Choose root over ratio by looking at the exponent
  • Recognise the inconclusive case immediately

Transcript

Everything said in the video, section by section (1,455 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 7.6

The Root Test

The Ratio Test compared each term to the one before it. The Root Test does something related but different. It takes the nth root of the term. And the reason that helps is simple. When the entire term is raised to the nth power — which happens more often than you'd expect — taking the nth root cancels that exponent outright. What's left is usually something you can evaluate at a glance. So the two tests are close cousins. But they're tuned for different structures.

The Test

Take the nth Root Instead of a Ratio

Here's the theorem. For a series, let L be the limit of the absolute value of a n, raised to the one over n. Then three cases. If L is less than one, the series converges absolutely. If L is greater than one, or infinite, the series diverges. And if L equals one, the test yields no information. Use a different test. It helps to picture those outcomes on a number line. Everything strictly left of one gives absolute convergence. Everything strictly right of one — including L infinite — gives divergence. And the single point at L equals one gives nothing at all. So reach for the Root Test when the entire term is raised to the nth power. That's its natural home.

Why

Another Geometric Comparison

The justification rests entirely on comparison with a geometric series. Just like the Ratio Test. And it's worth seeing, because it explains the role of the number one. Suppose L is less than one. Choose a fixed number r strictly between L and one. Since the nth roots approach L, and r sits above L, all but finitely many of those roots must fall below r. So there's an index N such that the nth root of absolute a n is less than r, for every n at least N. Now raise both sides to the nth power. That's legitimate, because both sides are nonnegative and the function t to the n is increasing on the nonnegative reals. You get absolute a n less than r to the n, for every n at least N. And the sum of r to the n is geometric, with ratio r between zero and one. So it converges. By the Direct Comparison Test, the sum of absolute a n converges as well. And absolute convergence implies convergence. Now suppose L is greater than one. Then the nth root of absolute a n exceeds one for all sufficiently large n. Which means absolute a n itself exceeds one for those n. So the terms fail to approach zero, and the nth Term Test applies immediately. The conclusion in that case is stronger than mere divergence. The terms don't even shrink.

The Computation

Two Standard Limits Do Most of the Work

In practice you apply the Root Test by writing down the nth root, simplifying it algebraically before taking any limit, and only then evaluating. And two algebraic facts do nearly all of the work. First. For any fixed positive constant c, the limit of c to the one over n is one. Second. For any fixed real p, the limit of n to the p, all to the one over n, is also one. In particular, n to the one over n approaches one. The consequence is important. Constant factors and polynomial factors are invisible to the root test. They contribute a factor of one to L, and can be discarded once the limit is taken. Let's do two. First, the sum of two n over three n plus four, all to the n. All terms are positive, so the absolute value changes nothing. Take the nth root. The exponent n and the exponent one over n multiply to give one. So the outer power simply cancels. You're left with two n over three n plus four. Divide numerator and denominator by n. You get two over three plus four over n. As n goes to infinity, that's two thirds. About zero point six six seven. Less than one, so the series converges absolutely. Second. The sum of five n squared plus one, over two n squared plus three n, all to the n. Again the nth root cancels the outer exponent. You're left with the rational function. Divide by n squared. The limit is five halves. Two point five. Greater than one, so the series diverges. And the reasoning is worth restating. Since the nth roots eventually exceed one, the terms themselves eventually exceed one. So they cannot approach zero.

Worked Examples 3 and 4

What Washes Out, and What Survives

Two more examples, showing what washes out and what survives. Third. The sum of seven n cubed over four to the n. Take the nth root of the absolute value, and distribute it across the product and quotient — using the identity that the nth root of a product is the product of the nth roots. You get seven to the one over n, times n cubed to the one over n, all over four. Now apply the two standard limits. Seven to the one over n goes to one. And n cubed to the one over n also goes to one. So L is one times one, over four. One quarter. Less than one, so the series converges absolutely. Notice that the constant seven and the polynomial n cubed both washed out completely. Only the exponential base survived. Fourth. The sum of one plus one over n, raised to the minus n squared. This term would be punishing under the Ratio Test. But watch what the root does. The nth root multiplies the exponent minus n squared by one over n. Reducing it to minus n. In a single stroke. So you're left with one plus one over n, to the minus n. Now recall the definition of the natural base. One plus one over n, to the n, approaches e. So its reciprocal approaches one over e. About zero point three six eight. Less than one. The series converges absolutely. Now compare what the Ratio Test would have required. You'd have had to simplify one plus one over n plus one, to the minus n plus one squared — divided by one plus one over n, to the minus n squared. An expression with no obvious cancellation at all. So structure, not difficulty of the underlying series, dictates the choice of test.

Choosing a Test

Match the Structure to the Tool

Let's make that structural matching explicit. If the exponent is a function of n — like n squared — use the Root Test. Because the root reduces n squared to n. If the term contains a factorial, use the Ratio Test. Factorials cancel cleanly in a ratio. Not in a root. If the term is a rational function of n only, with no exponent involving n, use a Comparison Test or the Integral Test. Because both the root and the ratio will give L equals one and stall. And if you have a constant base raised to the n, times a polynomial, either test works. The root is usually faster. Now let's see the inconclusive case explicitly. Apply the Root Test to the harmonic series, sum of one over n. By the second standard limit with p equals minus one, L is one over the limit of n to the one over n. Which is one over one. One. Now apply it to the sum of one over n squared. The same property gives L equals one over the limit of n to the one over n, squared. Which is one over one squared. Also one. Both computations return L equal to one. Yet the harmonic series is the standard example of a divergent series, while the p-series with p equal to two converges. That pair is the cleanest possible demonstration that L equals one carries no information. Two series with opposite behavior producing identical values of L. And the p-Series Test settles both cases at once.

Key Takeaways

What to Carry Forward

Five takeaways. Compute L, the limit of absolute a n to the one over n. L less than one gives absolute convergence. L greater than one, or infinite, gives divergence. L equal to one gives nothing. Like the Ratio Test, it's a geometric comparison in disguise. L less than one traps the terms beneath r to the n, for some r less than one. Simplify the nth root algebraically before taking any limit. And the two standard limits do most of the work. Constants and polynomial factors are invisible to the Root Test. Only exponential structure survives. And match the tool to the structure. Reach for the root when the exponent involves n. Reach for the ratio when factorials appear. And reach for neither when the term is a rational function of n alone. Next up is section seven point seven — Strategy for Testing Series. Which organizes every test in this chapter into a single decision procedure.

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FAQ

Questions students ask us

About The Root Test

What is the Root Test for series?
The Root Test decides convergence by computing L = \lim_{n\to\infty} |a_n|^{1/n}. If L < 1, the series \sum a_n converges absolutely. If L > 1 or L = \infty, it diverges. If L = 1, the test gives no information. It works because L < 1 forces |a_n| < r^n for some r < 1 and all large n, which is a convergent geometric series.
When should you use the Root Test instead of the Ratio Test?
Use the Root Test when the whole term is raised to a power involving n, such as \left(\frac{2n}{3n+4}\right)^n or (1+1/n)^{-n^2}, because the nth root cancels or reduces that exponent. Prefer the Ratio Test when factorials appear, since they cancel cleanly in a ratio. For a rational function of n alone, both tests give L = 1, so use a comparison test or the Integral Test instead.
How do you use the Root Test on \sum \frac{7n^3}{4^n}?
Take the nth root, simplify, and apply two standard limits to get L = 1/4, so the series converges absolutely. The root splits as \frac{7^{1/n}(n^3)^{1/n}}{4}. Since c^{1/n} \to 1 for any positive constant c and (n^p)^{1/n} \to 1 for any real p, both factors in the numerator approach 1. Constants and polynomial factors wash out, and only the exponential base survives.
What does it mean when the Root Test gives L = 1?
It means the test is inconclusive, so you must switch to a different test. For example, \sum 1/n and \sum 1/n^2 both give L = 1 because n^{1/n} \to 1. Yet the harmonic series diverges, while the p-series with p = 2 converges. A common mistake is reading L = 1 as a verdict. It says nothing either way, and here the p-Series Test settles both cases.

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