What Is The Radius Of Convergence?
The radius of convergence of a power series is the distance from the series' center out to where it stops converging. For a power series written as
$$\sum_{n=0}^{\infty} c_n (x-a)^n,$$
the number $a$ is the center and each $c_n$ is a coefficient. The radius of convergence $R$ is the single number for which the series converges for every $x$ with $|x-a| < R$ and diverges for every $x$ with $|x-a| > R$.
There are exactly three cases, and no fourth is possible:
$R = 0$. The series converges only at the center $x = a$ and nowhere else.
$R = \infty$. The series converges for every real number $x$.
$0 < R < \infty$. The series converges on an interval around $a$ and diverges outside it.
This three-way split is a theorem, not a coincidence. A power series can never converge on, say, a scattered set of points or on a one-sided ray. It always locks onto a symmetric interval around its center, which is why "radius" is the right word: the set of points where the series behaves is a segment of half-width $R$ centered at $a$. The idea sits inside the broader study of the convergence and divergence of series.
How Do You Find The Radius Of Convergence?
The standard tool is the ratio test applied to the terms of the series. Take the absolute value of the ratio of consecutive terms and send $n$ to infinity:
$$\lim_{n \to \infty}\left|\frac{c_{n+1}(x-a)^{n+1}}{c_n (x-a)^n}\right| = |x-a|\lim_{n \to \infty}\left|\frac{c_{n+1}}{c_n}\right|.$$
The ratio test says the series converges when this limit is below $1$. Setting $|x-a|\cdot L < 1$, where $L = \lim_{n \to \infty}\left|\dfrac{c_{n+1}}{c_n}\right|$, gives $|x-a| < 1/L$. Reading off the half-width:
$$R = \frac{1}{L} = \lim_{n \to \infty}\left|\frac{c_n}{c_{n+1}}\right|.$$
The geometric meaning sits right beside the algebra. On the number line, $|x-a| < R$ is the open segment stretching a distance $R$ to the left and right of the center. Everything inside that segment converges; everything outside diverges. In the complex plane the same rule draws a disk of radius $R$, which is where the name "radius" comes from.
When the coefficients carry an $n$-th power rather than a factorial or a ratio that simplifies, the root test is often cleaner:
$$\frac{1}{R} = \lim_{n \to \infty}\sqrt[n]{|c_n|}.$$
Both tests return the same $R$. Choose the ratio test when consecutive coefficients divide neatly, and the root test when the coefficient is something like $c_n = (2n)^n$ where an $n$-th root collapses the power.
What Are Some Worked Radius Of Convergence Examples?
Each example applies the ratio test and states $R$. The interval, with its endpoint checks, comes in the next section.
Example 1: The series $\sum_{n=1}^{\infty} \dfrac{x^n}{n}$.
Here $c_n = \dfrac{1}{n}$ and the center is $a = 0$. Compute the coefficient ratio:
$$R = \lim_{n \to \infty}\left|\frac{c_n}{c_{n+1}}\right| = \lim_{n \to \infty}\frac{1/n}{1/(n+1)} = \lim_{n \to \infty}\frac{n+1}{n} = 1.$$
Final answer: $R = 1$.
Example 2: The exponential series $\sum_{n=0}^{\infty} \dfrac{x^n}{n\text{ factorial}}$.
Here $c_n = \dfrac{1}{n\text{ factorial}}$. The key step is the factorial ratio $\dfrac{(n+1)\text{ factorial}}{n\text{ factorial}} = n+1$:
$$R = \lim_{n \to \infty}\left|\frac{c_n}{c_{n+1}}\right| = \lim_{n \to \infty}\frac{1/(n\text{ factorial})}{1/((n+1)\text{ factorial})} = \lim_{n \to \infty}(n+1) = \infty.$$
Final answer: $R = \infty$. This series converges for every $x$, and it happens to equal $e^x = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6} + \dfrac{x^4}{24} + \cdots$, which is why $e^x$ can be evaluated at any input. It is the Maclaurin series of the exponential function.
Example 3: The series $\sum_{n=0}^{\infty} n\text{ factorial},; x^n$.
Here $c_n = n\text{ factorial}$, so the ratio flips:
$$R = \lim_{n \to \infty}\left|\frac{c_n}{c_{n+1}}\right| = \lim_{n \to \infty}\frac{n\text{ factorial}}{(n+1)\text{ factorial}} = \lim_{n \to \infty}\frac{1}{n+1} = 0.$$
Final answer: $R = 0$. The factorials grow so fast that the terms blow up for any $x \neq 0$, so the series converges only at its center.
Example 4: A shifted series $\sum_{n=0}^{\infty} \dfrac{(x-2)^n}{3^n}$.
Here $c_n = \dfrac{1}{3^n}$ and the center is $a = 2$. The coefficient ratio is constant:
$$R = \lim_{n \to \infty}\left|\frac{c_n}{c_{n+1}}\right| = \lim_{n \to \infty}\frac{1/3^n}{1/3^{n+1}} = \lim_{n \to \infty} 3 = 3.$$
Final answer: $R = 3$, centered at $a = 2$.
What Is The Interval Of Convergence, And How Do You Find It?
The interval of convergence is the full set of $x$-values where the series converges. It always contains the open interval $(a-R,, a+R)$, and it may or may not include the two endpoints $x = a-R$ and $x = a+R$. The ratio test is silent at the endpoints, because there the limit equals exactly $1$ and the test is inconclusive. Each endpoint has to be substituted back into the series and tested on its own.
The workflow is three steps:
Find $R$ with the ratio test.
Write the open interval $(a-R,, a+R)$.
Substitute each endpoint into the original series and test the resulting numeric series separately.
Interval for Example 1, $\sum \dfrac{x^n}{n}$. Here $R = 1$ and $a = 0$, so the open interval is $(-1, 1)$. Now test the two endpoints:
At $x = 1$ the series becomes $\sum_{n=1}^{\infty}\dfrac{1}{n}$, the harmonic series, which diverges.
At $x = -1$ the series becomes $\sum_{n=1}^{\infty}\dfrac{(-1)^n}{n}$, the alternating harmonic series, which converges by the alternating series test.
One endpoint is in, the other is out, so the interval of convergence is $[-1, 1)$. Notice that the two endpoints, though the same distance from the center, behave differently. That is exactly why each must be checked by hand.
Interval for Example 4, $\sum \dfrac{(x-2)^n}{3^n}$. Here $R = 3$ and $a = 2$, so the open interval is $(2-3,, 2+3) = (-1, 5)$. Test the endpoints:
At $x = 5$ the series becomes $\sum \dfrac{3^n}{3^n} = \sum 1$, which diverges (the terms do not approach $0$).
At $x = -1$ the series becomes $\sum \dfrac{(-3)^n}{3^n} = \sum (-1)^n$, which also diverges (the terms oscillate and never settle).
Both endpoints fail, so the interval of convergence is the open interval $(-1, 5)$.
Table: The four worked series, their radius, and their interval of convergence.
Power series | Center $a$ | Radius $R$ | Interval of convergence |
|---|---|---|---|
$\sum \dfrac{x^n}{n}$ | $0$ | $1$ | $[-1, 1)$ |
$\sum \dfrac{x^n}{n\text{ factorial}}$ | $0$ | $\infty$ | $(-\infty, \infty)$ |
$\sum n\text{ factorial},; x^n$ | $0$ | $0$ | ${0}$ only |
$\sum \dfrac{(x-2)^n}{3^n}$ | $2$ | $3$ | $(-1, 5)$ |
The radius is one number; the interval is the radius plus the verdict at each end. Keeping those two ideas apart is the whole skill.
Why Does The Radius Of Convergence Work?
The three-case behavior looks almost too clean the first time, but it follows from how the terms $c_n(x-a)^n$ grow.
The power $(x-a)^n$ decides the race. For $x$ close to the center, $|x-a|$ is small and its $n$-th power shrinks fast enough to overcome the coefficients, so the terms head to zero and the series converges. Push $x$ far enough out and $|x-a|^n$ grows faster than the coefficients can tame, so the terms blow up and the series diverges. The tipping point between the two is a single distance, and that distance is $R$.
The disk explains the symmetry. A power series is really a function on the complex plane, and it converges on a disk centered at $a$. The radius $R$ is the distance from $a$ to the nearest point where the underlying function misbehaves (a singularity). Because a disk is symmetric, the real interval it cuts out is symmetric too, which is why convergence spreads the same distance in both directions.
The endpoints are the boundary circle. The two endpoints of the real interval sit exactly on the boundary of that disk, where the term $|x-a|^n = R^n$ is perfectly balanced against the coefficients. On this knife-edge the outcome depends on the fine detail of the coefficients, so it can go either way, and only a direct test settles it.
Seen through the disk, "radius of convergence" stops being jargon: it is the literal radius of the region where an infinite series built from powers actually adds up to a finite number.
Who Discovered The Radius Of Convergence?
The idea grew out of the drive to make infinite series rigorous in the 1800s, after a century of using them loosely.
One more name belongs beside them:
Niels Henrik Abel (1802–1829, Norway) studied what happens on the boundary circle itself. His theorem on the behavior of a series at an endpoint is the reason the endpoint check has firm footing, and it explains cases like the alternating harmonic series converging at $x = -1$ while the harmonic series diverges at $x = 1$.
Where Is The Radius Of Convergence Used In The Real World?
Power series are how many functions are actually computed and approximated, so their radius sets the limit of where those approximations can be trusted.
Physics and perturbation theory: solutions in quantum mechanics and celestial mechanics are often written as power series in a small parameter, and the radius of convergence tells physicists how strong that parameter can get before the expansion stops describing reality.
Numerical computing: a calculator or library evaluating $e^x$, $\sin x$, or $\ln(1+x)$ sums a power series, and it must know the radius to decide when the series is valid and when to switch methods.
Signal processing: the transfer functions of digital filters are power series in a complex variable, and the radius of convergence marks the region of stability for the filter.
Finance and economics: present-value and annuity formulas are geometric-type series, and their radius of convergence is the discount-rate range where the closed-form sum is legitimate.
Differential equations: power-series solutions built through a Taylor series are only usable inside their radius, which bounds how far from the starting point the solution can be extended.
One number, the radius, draws the line between an approximation you can rely on and one that has quietly stopped working.
What Are The Most Common Mistakes With The Radius Of Convergence?
These four errors account for most lost marks on this topic, and each matches a question real students ask on r/calculus, r/learnmath, and course exam-comment handouts.
Forgetting to test the endpoints.
Where it slips in:
A student finds $R = 1$ for $\sum \dfrac{x^n}{n}$, writes the interval as $(-1, 1)$, and stops.
Don't do this:
Do not assume the interval is open. The ratio test is inconclusive at the endpoints and says nothing about them.
The correct way:
Substitute each endpoint into the series and test it separately. For $\sum \dfrac{x^n}{n}$, $x = -1$ converges but $x = 1$ diverges, so the interval is $[-1, 1)$, not $(-1, 1)$.
Mishandling the factorial ratio.
Where it slips in:
On $\sum \dfrac{x^n}{n\text{ factorial}}$, a student cancels the factorials wrongly, treating $\dfrac{(n+1)\text{ factorial}}{n\text{ factorial}}$ as $n$ or as $1$.
Don't do this:
Do not guess the cancellation. The ratio of consecutive factorials is not the ratio of the numbers.
The correct way:
Use $(n+1)\text{ factorial} = (n+1)\cdot n\text{ factorial}$, so $\dfrac{(n+1)\text{ factorial}}{n\text{ factorial}} = n+1$. That gives $R = \lim (n+1) = \infty$, the correct answer.
Confusing the radius with the interval.
Where it slips in:
A student reports "$R = [-1, 1)$" or answers "the interval is $3$," blending the two separate objects.
Don't do this:
Do not use one as the other. The radius is a single non-negative number (or infinity); the interval is a set of $x$-values.
The correct way:
State them apart. For $\sum \dfrac{(x-2)^n}{3^n}$, write $R = 3$ (a number) and interval of convergence $(-1, 5)$ (a set centered at $a = 2$).
Inverting the ratio-test limit the wrong way.
Where it slips in:
A student computes the ratio-test limit $L$ and reports $R = L$ instead of $R = 1/L$, so a limit of $\tfrac{1}{3}$ is written as $R = \tfrac{1}{3}$ rather than $R = 3$.
Don't do this:
Do not confuse the limit of the term ratio with the radius. The convergence condition is $|x-a|\cdot L < 1$, which solves to $|x-a| < 1/L$.
The correct way:
Take the reciprocal: $R = 1/L$. Equivalently, work with $R = \lim\left|\dfrac{c_n}{c_{n+1}}\right|$ from the start, which already has the coefficients in the correct order.
Practice Problems On The Radius Of Convergence
Work each one, then check against the answer. Every answer is verified.
Find $R$ for $\sum_{n=1}^{\infty}\dfrac{x^n}{2^n}$.
(Answer: $c_n = 1/2^n$, so $R = \lim\dfrac{2^{n+1}}{2^n} = 2$.)Find $R$ for $\sum_{n=0}^{\infty} n,x^n$.
(Answer: $c_n = n$, so $R = \lim\dfrac{n}{n+1} = 1$.)Find the interval of convergence for $\sum_{n=1}^{\infty}\dfrac{x^n}{n^2}$.
(Answer: $R = 1$; at $x = 1$, $\sum 1/n^2$ converges, and at $x = -1$, $\sum (-1)^n/n^2$ converges absolutely, so the interval is $[-1, 1]$.)Find $R$ for $\sum_{n=0}^{\infty}\dfrac{x^n}{(2n)\text{ factorial}}$.
(Answer: $R = \infty$, since $\dfrac{(2n)\text{ factorial}}{(2n+2)\text{ factorial}} = \dfrac{1}{(2n+1)(2n+2)} \to 0$, so the reciprocal limit is $\infty$.)Find the interval of convergence for $\sum_{n=0}^{\infty}\dfrac{(x-3)^n}{4^n}$.
(Answer: $R = 4$, center $a = 3$, open interval $(-1, 7)$; both endpoints give $\sum (\pm 1)^n$, which diverge, so the interval is $(-1, 7)$.)Find $R$ for $\sum_{n=1}^{\infty}\dfrac{(-1)^n x^n}{\sqrt{n}}$.
(Answer: $c_n = (-1)^n/\sqrt{n}$, so $R = \lim\dfrac{\sqrt{n+1}}{\sqrt{n}} = 1$.)
Where Should You Go Next After The Radius Of Convergence?
The radius of convergence is one link in the chain of series tools, and several natural doors open from here.
Power series. Step back to the object itself, how power series are built, added, differentiated, and integrated term by term.
The ratio test. Sharpen the single tool that produces the radius, including the cases where the limit is inconclusive.
Taylor series. See where these power series come from, as expansions of functions whose radius decides how far the approximation is valid.
Binomial series. A worked family of power series whose radius of convergence is exactly $1$, a clean case to practice on.
If your child is meeting power series for the first time, a live Bhanzu trainer teaches the radius of convergence from the disk picture up, so the endpoint checks feel like part of one idea, in the Bhanzu math program.
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