The Ratio Test for Power Series

From Coefficient Ratios to Convergence

The Cauchy-Hadamard Formula determines the radius of a power series from the limsup of the coefficient roots. When consecutive coefficients have a regular ratio, there is often a more direct route: apply the Ratio Test to the terms of the series at a fixed input. The resulting test identifies which distances from the center give absolute convergence and which give divergence.

The input $x$ must be held fixed while applying the test. For a power series centered at $a$, the $n$th term is $c_n(x-a)^n$. At any fixed $x\neq a$, the ratio of consecutive term magnitudes combines the coefficient ratio with the fixed distance $|x-a|$. This is what turns information about the coefficients into a radius.

The Ratio Criterion for a Power Series

Suppose the coefficients are nonzero from some index onward, so their consecutive ratios are defined eventually. The ratio limit may be a nonnegative finite number or infinity. A finite positive limit gives a finite positive radius; the extreme limits correspond to an infinite or zero radius.

Theorem (Ratio Criterion for the Radius of a Power Series): Let $\sum_{n=0}^{\infty}c_n(x-a)^n$ be a power series, and suppose $c_n\neq0$ for all sufficiently large $n$. If $$ \lim_{n\to\infty}\frac{|c_{n+1}|}{|c_n|}=L, $$ where $L\in[0,\infty]$, then its radius of convergence is $R=1/L$, with the conventions $1/0=\infty$ and $1/\infty=0$. This criterion does not determine convergence at points where $|x-a|=R$.

Proof. Fix $x\neq a$, and write $u_n=c_n(x-a)^n$ for the $n$th term. For all sufficiently large $n$, $u_n\neq0$, and

\frac{|u_{n+1}|}{|u_n|} = \frac{|c_{n+1}|\,|x-a|^{n+1}}{|c_n|\,|x-a|^n} = \frac{|c_{n+1}|}{|c_n|}|x-a|.

If $0<L<\infty$, this ratio tends to $L|x-a|$. When $|x-a|<1/L$, the limit is less than $1$, so the Ratio Test gives absolute convergence. When $|x-a|>1/L$, the limit is greater than $1$, so the Ratio Test gives divergence. Thus the series converges absolutely at every point closer than $1/L$ to $a$, and diverges at every point farther away. Its radius is $1/L$.

If $L=0$, then for every fixed $x\neq a$ the term-ratio limit is $0$, which is less than $1$. The Ratio Test gives absolute convergence for every such $x$; the series also converges at $x=a$, where all terms except the constant term vanish. Its radius is infinite.

If $L=\infty$, then for every fixed $x\neq a$ the term-ratio limit is infinite, hence greater than $1$. The Ratio Test gives divergence at every such point. The series converges at its center, so its radius is zero. These cases establish the stated radius. At $|x-a|=1/L$ when $0<L<\infty$, the term-ratio limit is $1$, and the Ratio Test gives no conclusion; endpoint behavior must be checked separately. $\square$

The ratio criterion is consistent with the Cauchy-Hadamard Formula. Earlier in this course, the theorem “A Consecutive-Ratio Limit Determines the Root Limit” established that, under the corresponding coefficient hypotheses, a consecutive-ratio limit determines the coefficient-root limit. The ratio approach is especially convenient when the coefficient quotient simplifies more easily than the roots.

Worked Examples: Calculating the Radius

Worked Example: A Finite Positive Ratio Limit

Consider the power series

\sum_{n=0}^{\infty}\frac{n+1}{5^n}(x-2)^n.

Here $c_n=(n+1)/5^n$, which is nonzero for every $n$. For each $n\geq0$,

\frac{|c_{n+1}|}{|c_n|} = \frac{(n+2)/5^{n+1}}{(n+1)/5^n} = \frac{n+2}{5(n+1)} \longrightarrow \frac15.

The Ratio Criterion for the Radius gives $R=1/(1/5)=5$. Therefore the series converges absolutely when $|x-2|<5$, or $-3<x<7$, and diverges when $|x-2|>5$. At either boundary point, the $n$th term has magnitude

\frac{n+1}{5^n}5^n=n+1,

which does not tend to zero. By the Necessary Condition for Series Convergence, the series diverges at both boundary points as well. The radius calculation identifies the open interval; the term test supplies the endpoint conclusions in this example.

Worked Example: A Zero Ratio Limit

Consider

\sum_{n=0}^{\infty}\frac{2^n}{n!}(x+1)^n.

The coefficients are nonzero, and their consecutive ratios are

\frac{|c_{n+1}|}{|c_n|} = \frac{2^{n+1}/(n+1)!}{2^n/n!} = \frac{2}{n+1} \longrightarrow0.

The ratio criterion gives $R=\infty$. To see the test directly at any fixed $x$, the term-ratio limit is $0\cdot|x+1|=0$, which is less than $1$. Thus the series converges absolutely for every real $x$. This conclusion includes $x=-1$, the center, where the series reduces to its constant term.

Worked Example: An Infinite Ratio Limit

Now consider

\sum_{n=0}^{\infty}\frac{n!}{4^n}(x-3)^n.

Every coefficient is nonzero, and

\frac{|c_{n+1}|}{|c_n|} = \frac{(n+1)!/4^{n+1}}{n!/4^n} = \frac{n+1}{4} \longrightarrow\infty.

The radius is therefore $0$. For any fixed $x\neq3$, the ratio of consecutive term magnitudes tends to infinity, so the Ratio Test gives divergence. At $x=3$, every term of positive degree vanishes and the series converges to its constant term. Thus convergence at the center is the only convergence point.

Why the Boundary Still Needs Separate Attention

When $0<L<\infty$, substituting a boundary point into the term-ratio limit gives $L|x-a|=1$. The Ratio Test is inconclusive when the ratio limit equals $1$. That does not mean the series necessarily diverges, nor does it mean it necessarily converges: the terms may have different sizes and different signs even when their consecutive magnitude ratio tends to $1$.

Worked Example: Equal Radius, Different Endpoint Behavior

Consider the power series

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

Its coefficients satisfy

\frac{|c_{n+1}|}{|c_n|} = \frac{1/(n+1)}{1/n} = \frac{n}{n+1} \longrightarrow1.

The radius is $R=1$. At $x=2$, the series becomes $\sum_{n=1}^{\infty}1/n$, which diverges by the $p$-Series Convergence Criterion with $p=1$. At $x=0$, it becomes $\sum_{n=1}^{\infty}(-1)^n/n$. The magnitudes $1/n$ decrease to zero, so the Alternating Series Test gives convergence. It is not absolute convergence, because the series of absolute values is the divergent harmonic series. The same coefficient-ratio limit thus gives the radius but does not decide either endpoint on its own.

The Radius of Convergence theorem and the Endpoint Classification of the Interval of Convergence theorem provide the right framework: use the radius to settle points strictly inside or outside, then test any finite-radius endpoints individually. The Ratio Test is an efficient way to obtain the radius when the coefficient ratios have a limit, not a substitute for endpoint analysis.

When the Coefficient Ratios Do Not Apply

The ratio criterion has a hypothesis that is easy to overlook: the coefficients must be nonzero eventually. If the coefficients are eventually zero, a coefficient ratio may be undefined, but the power series is just a polynomial after finitely many terms. A polynomial converges for every real $x$, so its radius is infinite. This is a separate simple case, rather than an application of the ratio criterion.

More generally, the criterion may be unavailable if the coefficients have recurring zeros or if their consecutive ratios do not converge. That does not imply that the radius fails to exist. The Cauchy-Hadamard Formula uses the limsup of the coefficient roots and remains applicable in situations where a consecutive-ratio limit does not exist. The ratio method is powerful when its limit exists, but it is not a universal replacement for the root formula.

A final practical point is that the coefficient ratio is taken in absolute value. Signs do not affect the radius: the Ratio Test applied at a fixed $x$ concerns the magnitudes of consecutive terms. Signs may matter at boundary points, as the alternating endpoint above demonstrates, but they do not change the ratio-based radius calculation.

Takeaway: If the coefficients are eventually nonzero and $|c_{n+1}|/|c_n|\to L$, then the radius is $1/L$, with infinite and zero radii for $L=0$ and $L=\infty$. At a finite-radius boundary, the Ratio Test is inconclusive and each endpoint requires its own test.

Check Your Understanding

Use the coefficient-ratio criterion and the examples above to answer the following questions.

If $|c_{n+1}|/|c_n|\to 1/4$, what is the radius of the power series?
Why does a coefficient-ratio limit of zero imply convergence at every fixed real input?
For the series $\sum_{n=0}^{\infty}(n+1)(x-4)^n/3^n$, calculate the coefficient-ratio limit and radius.
What does the Ratio Test say at a point where the term-ratio limit is $1$?
How should the radius be determined if the coefficients are eventually zero?
Why can signs affect endpoint convergence without affecting the radius obtained from coefficient ratios?