AP Calculus BC

Interval of Convergence Finder: Ratio Test and Endpoints

Q: What is the difference between the radius and the interval of convergence?

The radius of convergence R is a single distance: the series converges for every x within R of the center c and diverges beyond it. The interval of convergence is the actual set of x values, and getting it takes one more step, testing the two endpoints x = c \pm R. The radius tells you the interval is (c - R,\, c + R) up to its edges; the endpoint tests decide whether each edge is included, so the interval can be open, half-open, or closed (Topic 10.13).

Q: Why can't the ratio test decide the endpoints?

At an endpoint x = c \pm R the distance |x - c| equals the radius R exactly, so the ratio test limit comes out to L = 1. That is the inconclusive case of the ratio test (Topic 10.8): L 1 proves divergence, but L = 1 proves nothing. So you substitute the endpoint value into the series to get a number series and apply a different test, usually the Alternating Series Test or the p-series rule.

Q: Can the interval of convergence be a single point or all real numbers?

Yes, both extremes happen. If R = \infty the series converges for every real number and the interval is (-\infty,\, \infty); the series \sum \frac{x^n}{n!} for e^x is the standard example. If R = 0 the series converges only at its center and the interval collapses to the single point \{c\}; \sum n! \cdot x^n does this, because the factorial forces the terms to blow up for any x \ne 0. Every power series converges at least at its center, so the interval is never empty.

Q: How do I know whether to use a bracket or a parenthesis at an endpoint?

Substitute the endpoint's x-value into the power series, which turns it into an ordinary number series, then test that series. If it converges, the endpoint is included and you use a square bracket; if it diverges, the endpoint is excluded and you use a parenthesis. For \sum \frac{x^n}{n}, the endpoint x = -1 gives the convergent alternating harmonic series (bracket), while x = 1 gives the divergent harmonic series (parenthesis), so the interval is [-1,\, 1).

Find an interval of convergence in two stages: run the ratio test and solve L < 1 for the radius R and an open interval centered at c, then test each endpoint separately, because the ratio test gives L = 1 there and settles nothing. The final interval can be open, half-open, or closed.

Series

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Find the interval of convergence

\sum_{n=1}^{\infty} \frac{x^n}{n}

1Set up the ratio test

\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^{n+1}}{n+1} \cdot \frac{n}{x^n}\right| = |x| \cdot \frac{n}{n+1}

This tool takes one power series at a time and walks it through the ratio test in five staged reveals, in the order you would write on the exam: assemble the term ratio, take its limit $L$ , solve $L < 1$ for the radius, test the endpoints, then draw the interval. Finding the radius and interval of convergence of a power series is Topic 10.13, and the engine that produces the radius is the ratio test from Topic 10.8.

For a power series $\sum a_n$ with terms $a_n = c_n (x - c)^n$ , the ratio test compares the size of one term to the last. The powers of $(x - c)$ pull a factor of $|x - c|$ out front, and whatever is left from the coefficients $c_n$ collapses to a number in the limit:

L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|, \qquad \text{the series converges when } L < 1

Because $L$ always comes out proportional to $|x - c|$ , the condition $L < 1$ rearranges into a distance from the center. That distance is the radius $R$ :

L < 1 \iff |x - c| < R

Solving that inequality gives the open interval $(c - R,\, c + R)$ . The tool labels this the open interval so far, because the two endpoints $x = c - R$ and $x = c + R$ are still undecided.

The ratio test is deliberately silent at the endpoints. Substitute $x = c \pm R$ and the distance $|x - c|$ equals $R$ exactly, so $L = 1$ , which is the one case the ratio test cannot settle (Topic 10.8):

x = c - R \;\text{or}\; x = c + R \;\Rightarrow\; L = 1 \quad (\text{inconclusive})

So the tool does the step a solver skips: it substitutes each endpoint value back into the series to get a concrete number series, then picks a named test for it. An endpoint that produces an alternating series usually converges by the Alternating Series Test (Topic 10.7); one that produces the harmonic series $\sum \frac{1}{n}$ diverges as a $p$ -series with $p = 1$ (Topic 10.5); and terms that do not shrink to $0$ fail the nth-term test (Topic 10.3). Because each endpoint is decided on its own, the interval can end up open, half-open, or closed.

One endpoint can converge while the other diverges

That is why the number line uses two kinds of dots. A filled dot means the endpoint converges and is included (a square bracket); an open dot means it diverges and is excluded (a parenthesis). The series $\sum \frac{x^n}{n}$ converges at $x = -1$ (alternating) but diverges at $x = 1$ (harmonic), so its interval is the half-open $[-1,\, 1)$ : one bracket, one parenthesis.

Take the series the tool centers away from the origin, $\sum_{n=1}^{\infty} \frac{(x-2)^n}{n \cdot 3^n}$ . Its center is $c = 2$ , the detail most students drop. Watch the radius come from the ratio test and the interval come from the endpoints. First assemble the term ratio and simplify:

\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(x-2)^{n+1}}{(n+1) \cdot 3^{n+1}} \cdot \frac{n \cdot 3^n}{(x-2)^n}\right| = \frac{|x-2|}{3} \cdot \frac{n}{n+1}

Take the limit as $n \to \infty$ ; the fraction $\frac{n}{n+1}$ tends to $1$ , leaving $L$ proportional to $|x - 2|$ :

L = \lim_{n \to \infty} \frac{|x-2|}{3} \cdot \frac{n}{n+1} = \frac{|x-2|}{3}

Set the limit below $1$ and read off the radius:

L < 1 \iff |x-2| < 3 \;\Rightarrow\; R = 3

A radius of $3$ about the center $c = 2$ gives the open interval $(2 - 3,\, 2 + 3) = (-1,\, 5)$ . Now test both endpoints by substituting each $x$ -value into the series:

x = -1: \quad \sum_{n=1}^{\infty} \frac{(-3)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} \quad (\text{converges})

x = 5: \quad \sum_{n=1}^{\infty} \frac{3^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{1}{n} \quad (\text{diverges})

At $x = -1$ the series is the alternating harmonic series, which converges by the Alternating Series Test, so the left endpoint is included. At $x = 5$ the series is the plain harmonic series, a $p$ -series with $p = 1$ , which diverges, so the right endpoint is excluded. One bracket, one parenthesis, giving the interval of convergence:

[-1,\, 5)

The radius $R$ can be a finite positive number, infinity, or zero, and the tool's five series include at least one of each kind (three with a finite radius, one with $R = \infty$ , one with $R = 0$ ), so you learn to read the shape from the terms. The number line renders all three: a segment between two endpoints, the whole line with arrowheads, or a single point at the center.

Radius	Converges for	Endpoints	Series in the tool
$R$ finite, $R > 0$	$x$ within $R$ of the center	test both, separately	$\sum \frac{x^n}{n}$ gives $[-1,\, 1)$
$R = \infty$	every real number	none to test	$\sum \frac{x^n}{n!}$ , the series for $e^x$
$R = 0$	only the center $x = c$	the interval is one point	$\sum n! \cdot x^n$ converges only at $0$

When only odd powers appear

The series $\sum \frac{(-1)^n x^{2n+1}}{2n+1}$ , the Maclaurin series for $\arctan x$ , has only odd powers, so the consecutive-term ratio tends to $x^2$ , not $|x|$ ; solving $x^2 < 1$ still gives $R = 1$ . Both endpoints turn into alternating series that converge, so the interval is the closed $[-1,\, 1]$ , and $x = 1$ recovers the Leibniz formula for $\frac{\pi}{4}$ .

How this tool teaches instead of solving

You assemble the term ratio and simplify it to the limit $L$ yourself; a compute-only solver just prints the radius. You solve $L < 1$ for the radius $R$ and write the open interval before deciding any endpoint, so the two-stage structure stays in front of you. You substitute each endpoint into the series and choose its own named test, which is the exact step that makes the interval open, closed, or half-open. The number line marks included endpoints with filled dots and excluded ones with open dots, so a bracket versus a parenthesis becomes something you can see. Five series cover the finite, infinite, and zero-radius cases, training you to recognize the regime instead of memorizing one interval.

Frequently asked questions

What is the difference between the radius and the interval of convergence?

The radius of convergence $R$ is a single distance: the series converges for every $x$ within $R$ of the center $c$ and diverges beyond it. The interval of convergence is the actual set of $x$ values, and getting it takes one more step, testing the two endpoints $x = c \pm R$ . The radius tells you the interval is $(c - R,\, c + R)$ up to its edges; the endpoint tests decide whether each edge is included, so the interval can be open, half-open, or closed (Topic 10.13).

Why can't the ratio test decide the endpoints?

At an endpoint $x = c \pm R$ the distance $|x - c|$ equals the radius $R$ exactly, so the ratio test limit comes out to $L = 1$ . That is the inconclusive case of the ratio test (Topic 10.8): $L < 1$ proves convergence and $L > 1$ proves divergence, but $L = 1$ proves nothing. So you substitute the endpoint value into the series to get a number series and apply a different test, usually the Alternating Series Test or the $p$ -series rule.

Can the interval of convergence be a single point or all real numbers?

Yes, both extremes happen. If $R = \infty$ the series converges for every real number and the interval is $(-\infty,\, \infty)$ ; the series $\sum \frac{x^n}{n!}$ for $e^x$ is the standard example. If $R = 0$ the series converges only at its center and the interval collapses to the single point $\{c\}$ ; $\sum n! \cdot x^n$ does this, because the factorial forces the terms to blow up for any $x \ne 0$ . Every power series converges at least at its center, so the interval is never empty.

How do I know whether to use a bracket or a parenthesis at an endpoint?

Substitute the endpoint's $x$ -value into the power series, which turns it into an ordinary number series, then test that series. If it converges, the endpoint is included and you use a square bracket; if it diverges, the endpoint is excluded and you use a parenthesis. For $\sum \frac{x^n}{n}$ , the endpoint $x = -1$ gives the convergent alternating harmonic series (bracket), while $x = 1$ gives the divergent harmonic series (parenthesis), so the interval is $[-1,\, 1)$ .

Is finding the interval of convergence on the AP Calculus BC exam?

Yes. It is Topic 10.13 in Unit 10, Infinite Sequences and Series, which is BC-only and weighted 15 to 20 percent of the BC exam. Expect to find a radius with the ratio test and then justify each endpoint with a named test; the endpoint work is where most of the points live, since the ratio test alone only gets you the open interval.