The Root Test for Power Series

From Coefficient Growth to Convergence

The Ratio Test for Power Series used consecutive coefficient ratios to determine a radius of convergence. That approach is convenient when the ratios have a limit, but the coefficients need not have regular consecutive behavior. The Root Test gives another route: it measures the long-term size of each coefficient through its \(n\)th root, allowing irregular sequences to be handled by a limsup.

As in the previous tutorial, fix the input \(x\) before testing the series. For a power series centered at \(a\), the \(n\)th term is \(c_n(x-a)^n\). Its \(n\)th root in absolute value separates into a coefficient factor and a fixed distance factor. This separation is the key calculation: the coefficient roots describe the radius, while the distance from \(x\) to the center determines which side of the Root Test threshold the terms fall on.

The Root Criterion at a Fixed Input

For \(n\geq1\), set \(q_n=|c_n|^{1/n}\), and define the coefficient-root growth rate by \(L=\limsup_{n\to\infty}q_n\). The value \(L\) may be finite or infinite. The constant coefficient \(c_0\) does not affect convergence, so it does not enter this definition.

Theorem (Root Criterion for a Power Series): Let \(\sum_{n=0}^{\infty}c_n(x-a)^n\) be a power series and let \(L=\limsup_{n\to\infty}|c_n|^{1/n}\). For a fixed \(x\neq a\), put \(d=|x-a|\). If \(dL<1\), the series converges absolutely at \(x\). If \(dL>1\), the series diverges at \(x\). The criterion gives no conclusion when \(dL=1\). Here \(d\cdot\infty=\infty\) for \(d>0\).

Proof. Write \(u_n=c_n(x-a)^n\) for \(n\geq1\). Since \(d=|x-a|\) is fixed and positive,

|u_n|^{1/n} = \bigl(|c_n|d^n\bigr)^{1/n} = |c_n|^{1/n}d.

Therefore \(\limsup_{n\to\infty}|u_n|^{1/n}=dL\), with the same identity when \(L=\infty\). If \(dL<1\), the Root Test for series gives absolute convergence. If \(dL>1\), that test gives divergence. When \(dL=1\), the Root Test is inconclusive, so this calculation alone establishes neither convergence nor divergence. At \(x=a\), the power series converges directly: every term of positive degree vanishes, leaving \(c_0\). \(\square\)

This theorem is the fixed-input form of the Root Test. In terms of the radius, its strict inequalities say that the series converges absolutely when \(d<1/L\) and diverges when \(d>1/L\), using the conventions \(1/0=\infty\) and \(1/\infty=0\). This agrees with the Cauchy-Hadamard Formula established earlier in the course; that formula already identifies the radius as \(1/L\). The point here is to see directly how the Root Test produces the inside-and-outside classification, including when the coefficient roots themselves do not converge.

Worked Examples: Using Coefficient Roots

Worked Example: A Finite Root Growth Rate

Consider the power series centered at \(a=1\):

\sum_{n=0}^{\infty}\frac{n+1}{4^n}(x-1)^n.

For \(n\geq1\), its coefficient roots are

|c_n|^{1/n} = \left(\frac{n+1}{4^n}\right)^{1/n} = \frac{(n+1)^{1/n}}{4} \longrightarrow\frac14.

Thus \(L=1/4\), and the Root Criterion gives absolute convergence when \(|x-1|/4<1\), or \(|x-1|<4\). It gives divergence when \(|x-1|>4\). The finite radius is \(4\). At either boundary, \(|x-1|=4\), the product \(dL\) equals \(1\), so the Root Test does not decide convergence. That requires examining the endpoint series itself.

Worked Example: Coefficients with Recurring Zeros

Define \(c_0=0\), and for \(n\geq1\) let \(c_n=2^n\) when \(n\) is a perfect square and \(c_n=0\) otherwise. Consider \(\sum_{n=0}^{\infty}c_n(x-a)^n\). The coefficient roots are \(2\) at square indices and \(0\) at all other positive indices. Since there are infinitely many squares and infinitely many nonsquares,

\limsup_{n\to\infty}|c_n|^{1/n}=2.

For a fixed \(x\neq a\), write \(d=|x-a|\). If \(d<1/2\), then \(2d<1\), so the series converges absolutely by the Root Criterion. If \(d>1/2\), its terms at the square indices \(n=k^2\) have magnitudes

|c_{k^2}(x-a)^{k^2}| = 2^{k^2}d^{k^2} = (2d)^{k^2}.

Because \(2d>1\), these magnitudes do not tend to zero; in fact, they grow without bound. The terms of the series therefore fail the Necessary Condition for Series Convergence, so the series diverges. This example shows why a limsup is useful: the coefficient roots have no limit, but their limsup still gives the strict inside-and-outside classification.

Worked Example: A Boundary the Root Test Cannot Decide

Consider the series centered at zero

\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt n}x^n.

The absolute values of the coefficients satisfy

|c_n|^{1/n} = \left(\frac{1}{\sqrt n}\right)^{1/n} = n^{-1/(2n)} \longrightarrow1.

Consequently, \(L=1\). The series converges absolutely for \(|x|<1\) and diverges for \(|x|>1\), but the Root Test is inconclusive at both \(x=1\) and \(x=-1\). At \(x=1\), it becomes \(\sum_{n=1}^{\infty}(-1)^{n-1}/\sqrt n\). The magnitudes \(1/\sqrt n\) decrease to zero, so the Alternating Series Test gives convergence. At \(x=-1\), each term is \(-1/\sqrt n\), since \((-1)^{n-1}(-1)^n=-1\). The resulting series diverges by the \(p\)-Series Convergence Criterion with \(p=1/2\). Thus the Root Test identifies the radius but does not decide either endpoint.

A Geometric Tail Bound Inside the Radius

The Root Test can give more than a yes-or-no conclusion. If we stay within a fixed distance \(r\) of the center, strictly inside the radius, then the terms can be bounded by a geometric sequence uniformly over all such inputs. This is useful for controlling tails without recalculating the series separately at each point.

Theorem (Geometric Tail Bound on a Smaller Interval): Let \(L=\limsup_{n\to\infty}|c_n|^{1/n}\), and let \(r>0\) satisfy \(rL<1\). Choose \(q\) with \(rL<q<1\), interpreting the inequality in the usual way when \(L=0\). Then there is an integer \(N\geq1\) such that for every \(n\geq N\) and every \(x\) with \(|x-a|\leq r\),

|c_n(x-a)^n|\leq q^n.

In particular, for every integer \(M\geq N\), every such \(x\), and every \(K\geq M\),

\sum_{n=M}^{K}|c_n(x-a)^n| \leq \frac{q^M}{1-q}.

Proof. Since \(q/r>L\), the definition of limsup implies that there is an index \(N\) such that \(|c_n|^{1/n}<q/r\) for every \(n\geq N\). This also holds when \(L=0\), because \(q/r>0\). For any \(x\) with \(|x-a|\leq r\), it follows that

|c_n(x-a)^n| = |c_n|\,|x-a|^n \leq |c_n|r^n < \left(\frac{q}{r}\right)^n r^n = q^n

for every \(n\geq N\). Summing these inequalities from \(M\) to \(K\) and using the finite geometric-sum identity gives

\sum_{n=M}^{K}|c_n(x-a)^n| \leq \sum_{n=M}^{K}q^n = \frac{q^M(1-q^{K-M+1})}{1-q} \leq \frac{q^M}{1-q},

because \(0<q<1\) implies \(0\leq q^{K-M+1}\leq1\). The bound is independent of \(x\) in the stated interval and of the finite cutoff \(K\), as required. \(\square\)

For instance, if \(L=1/3\), then any \(r<3\) satisfies \(rL<1\). For a chosen such \(r\), one can select \(q\) strictly between \(r/3\) and \(1\). The theorem then bounds all sufficiently late terms throughout \(|x-a|\leq r\) by \(q^n\), and bounds every tail by a geometric tail. The condition that the interval be strictly inside the radius matters: at the boundary \(rL=1\), there is no \(q\) with \(rL<q<1\).

How to Read the Boundary Case

When \(dL=1\), the Root Test is inconclusive, not a verdict of divergence. It is also not a verdict of convergence. At such a point the \(n\)th roots of the term magnitudes have limsup \(1\), but this alone does not reveal how quickly the terms shrink, whether their signs alternate, or whether the terms even tend to zero. The worked endpoint example demonstrates that two boundary inputs can behave differently for the same power series.

A common error is to replace the coefficient-root limsup by a limit without checking that the limit exists. The square-index example had recurring zeros and no coefficient-root limit, yet its limsup still gave the needed classification. Another error is to use the strict convergence or divergence conclusions at equality. If \(dL=1\), return to the endpoint series and use an appropriate test, such as the Necessary Condition, comparison, or the Alternating Series Test.

The radius framework from earlier tutorials remains in force: the center always gives convergence; points strictly inside a finite positive radius give absolute convergence; points strictly outside diverge; and boundary points require separate analysis. The Root Test supplies a direct way to reach the strict conclusions from coefficient growth. The geometric tail bound adds a quantitative estimate whenever an entire interval stays strictly inside the radius.

Takeaway: For a fixed \(x\neq a\), the limsup of the \(n\)th roots of the term magnitudes is \(|x-a|L\), where \(L=\limsup |c_n|^{1/n}\). Values below \(1\) give absolute convergence and values above \(1\) give divergence; equality leaves the endpoint undecided. Strictly inside the radius, coefficient roots also yield geometric tail bounds.

Check Your Understanding

Use the Root Criterion and the geometric tail bound to answer the following questions.

If \(\limsup |c_n|^{1/n}=1/5\), at which distances from the center does the Root Criterion give absolute convergence, and at which distances does it give divergence?
Why can the coefficient-root limsup be useful when the coefficient roots do not have a limit?
For coefficients \(c_n=1/n!\), explain why their \(n\)th roots tend to zero and state what the Root Criterion implies about the radius.
What information does the Root Test provide when \(|x-a|L=1\)?
In the geometric tail bound, why must \(rL<1\) rather than \(rL\leq1\)?
For the square-index coefficient example, what happens to the term magnitudes along square indices when \(2|x-a|>1\)?

What You'll Learn