When the Ratio Test Gives No Radius

When Consecutive Ratios Do Not Settle

A power series may have coefficients whose consecutive ratios do not approach a limit. Some coefficients may be zero, or the ratios may fluctuate widely. In these cases, the ratio test may fail to provide a radius of convergence. That failure does not mean the radius cannot be found: the nth-root test measures coefficient growth without requiring consecutive ratios to settle.

For a power series centered at $c$, the relevant quantity is the limit superior of the nth roots of the coefficient magnitudes. This method works even when the coefficient sequence is irregular or has many zeros. We will use it to analyze both fluctuating coefficients and coefficients supported only at square indices.

The Root-Test View of a Power Series

Consider the power series $\sum_{n=0}^{\infty}a_n(x-c)^n$. At a fixed $x$, its $n$th term has magnitude $|a_n||x-c|^n$. For $x\ne c$, taking nth roots separates the coefficient growth from the distance to the center:

\bigl(|a_n||x-c|^n\bigr)^{1/n} = |a_n|^{1/n}|x-c|.

The root test says that a series converges absolutely if the limit superior of the nth roots of its term magnitudes is less than $1$, and diverges if that limit superior is greater than $1$. If it equals $1$, the root test gives no conclusion.

Definition: For the coefficients of a power series, set $$ L=\limsup_{n\to\infty}|a_n|^{1/n}, $$ where $L$ may be $0$, a positive finite number, or $+\infty$. The Cauchy–Hadamard formula defines the radius of convergence by $$ R=\frac{1}{L}, $$ with the conventions $1/0=+\infty$ and $1/(+\infty)=0$.

The limit superior records the largest growth rate that persists along arbitrarily large indices. It is particularly useful when some nth roots are small or zero, while others remain large. The formula gives the radius, but it does not by itself decide convergence at points exactly $R$ units from the center.

Why the Formula Works

For a fixed $x\ne c$, the root-test quantity for the series terms is the coefficient-root quantity multiplied by the fixed positive number $|x-c|$. Thus its limit superior is $L|x-c|$, interpreted as $+\infty$ if $L=+\infty$. The root test then identifies the points strictly inside and strictly outside the radius.

Take nth roots of the term magnitudes.
For $x\ne c$, the result is $|a_n|^{1/n}|x-c|$.

Take the limit superior.
Because $|x-c|$ is fixed and positive, the result is $L|x-c|$.

Apply the root test.
If $L|x-c|<1$, the series converges absolutely. If $L|x-c|>1$, it diverges.

When $0<L<+\infty$, the inequality $L|x-c|<1$ is equivalent to $|x-c|<1/L$, and $L|x-c|>1$ is equivalent to $|x-c|>1/L$. If $L=0$, every $x\ne c$ has root-test quantity $0$, so the series converges absolutely there and $R=+\infty$. If $L=+\infty$, every $x\ne c$ has root-test quantity $+\infty$, so the series diverges there and $R=0$.

Theorem: For $x\ne c$, the power series converges absolutely when $L|x-c|<1$, and diverges when $L|x-c|>1$, with the extended-value cases interpreted as above. At the center $x=c$, the series converges to $a_0$, since every term with $n\ge1$ is zero. The root-test inequalities give no conclusion when $L|x-c|=1$.

The separate center statement matters when $L=+\infty$. At $x=c$, the expression $L|x-c|$ would be $+\infty\cdot0$, which is undefined. The series nevertheless converges there directly: its zeroth term is $a_0$, and its remaining terms vanish.

Worked Examples

Worked Example: Ratios That Alternate Between Large and Small

Let $a_n=1$ for even $n$, and $a_n=4^n$ for odd $n$. Find the radius of convergence of $\sum_{n=0}^{\infty}a_n(x-c)^n$.

For even $n$, $|a_n|^{1/n}=1^{1/n}=1$. For odd $n$, $|a_n|^{1/n}=(4^n)^{1/n}=4$. Both kinds of indices occur arbitrarily far out, so the limit superior is $L=4$. Therefore,

R=\frac{1}{L}=\frac14.

The consecutive coefficient ratios do not settle. If $n$ is even, then $a_{n+1}/a_n=4^{n+1}$, which grows without bound. If $n$ is odd, then $a_{n+1}/a_n=1/4^n$, which tends to zero along odd indices. As a check of the radius, at $|x-c|=1/8$ the root-test quantity is $4(1/8)=1/2<1$, while at $|x-c|=1/2$ it is $4(1/2)=2>1$. These are respectively inside and outside the radius $1/4$.

Worked Example: Coefficients Supported Only at Squares

Let $a_n=1$ when $n=k^2$ for some positive integer $k$, and let $a_n=0$ otherwise. Take $a_0=0$. Find the radius of $\sum_{n=0}^{\infty}a_n(x-c)^n$.

At square indices $n=k^2$, the coefficient root is $1^{1/n}=1$. At all other positive indices it is $0^{1/n}=0$. There are arbitrarily large square indices, so the sequence of coefficient roots has limit superior $L=1$. Hence

R=\frac{1}{1}=1.

The consecutive ratio $a_{n+1}/a_n$ is not usable across all indices: when $a_n=0$, the quotient is undefined. Thus the ratio test cannot supply a radius through those coefficient ratios, but the nth-root calculation does. As a check, for $|x-c|=1/2$ the root-test quantity is $1/2<1$, and for $|x-c|=2$ it is $2>1$. The radius is therefore $1$.

Worked Example: Nonzero Coefficients with Irregular Ratios

Define $a_n=1$ when $n$ is a perfect square, and $a_n=2^n$ otherwise. Find the radius of $\sum_{n=0}^{\infty}a_n(x-c)^n$, taking $0$ to be a square.

At square indices, $|a_n|^{1/n}=1$. At nonsquare indices, $|a_n|^{1/n}=(2^n)^{1/n}=2$. There are arbitrarily large nonsquares, so $L=2$, and

R=\frac12.

Here all coefficients are nonzero, but their ratios still do not converge. At a square index $n$, the next index is not a square for all sufficiently large $n$, so $a_{n+1}/a_n=2^{n+1}$, which becomes arbitrarily large. At $n=k^2-1$, for $k\ge2$, the ratio is $a_{k^2}/a_{k^2-1}=1/2^{k^2-1}$, which tends to zero. Thus the ratios have no limit. Checking the radius with the root test, at $|x-c|=1/4$ the quantity is $2(1/4)=1/2<1$, while at $|x-c|=1$ it is $2>1$.

A Common Mistake: Treating Every Ratio Failure as a Radius

A frequent error is to use a ratio formula even when the ratio limit does not exist. For the square-supported coefficients above, one might try to write $a_{n+1}/a_n$ at every index and conclude that the radius is “the reciprocal of the ratio limit.” But the quotient is undefined whenever $a_n=0$, and there is no ratio limit to insert into such a formula. The correction is to calculate $\limsup |a_n|^{1/n}$ directly and use the Cauchy–Hadamard formula.

Another error is to treat the root-test equality as a convergence result. If $L|x-c|=1$, the root test is inconclusive; it does not say that the series converges or diverges. The radius identifies the boundary to investigate, not the outcome there.

Why This Method Matters

The ratio test is efficient when consecutive ratios approach a suitable limit, but coefficient sequences need not behave smoothly from one index to the next. Zeros, alternating growth rates, and other irregular patterns can prevent a ratio limit from existing. The root-based formula looks at the long-run exponential scale instead, so it remains informative in these situations.

For practical work, first inspect the coefficients. If their consecutive ratios settle, the ratio method may be convenient. If they do not, compute the nth roots and their limit superior. Keep the center separate when applying the root-test inequalities, and remember that the radius alone does not settle convergence at the boundary.

Check Your Understanding

For each question, use coefficient-root growth rather than assuming that a consecutive ratio limit exists.

If $|a_n|^{1/n}$ equals $1$ for even $n$ and $5$ for odd $n$, what are $L$ and the radius?
For coefficients equal to $1$ at square indices and $0$ elsewhere, why can the nth-root method be applied even though some consecutive coefficient ratios are undefined?
What does the root test conclude when $L|x-c|=1$?
Why must $x=c$ be handled separately if $L=+\infty$?
If $L=0$, what radius does the Cauchy–Hadamard formula assign?