Radius of Convergence Calculator

What is the radius of convergence?

A power series is a sum of the form

a₀ + a₁(x - c) + a₂(x - c)² + ...

The radius of convergence is the number R that tells you how far from the center c the series is guaranteed to converge. The series converges for every x inside the interval (c - R, c + R), and it diverges outside the closed interval [c - R, c + R]. Everything interesting happens at the two endpoints, where the series may converge or diverge depending on the term a_n.

The simplest example is the geometric series with a_n = 1 and center c = 0:

1 + x + x² + x³ + ...

This series converges to 1 / (1 - x) exactly when |x| < 1, so its radius of convergence is R = 1. Plug in x = 0.5 and the partial sums clearly approach 2. Plug in x = 1.5 and the terms grow without bound. The radius is the boundary between those two behaviors, and it lives entirely in the coefficients a_n: the same R applies no matter which x you pick, because it measures the coefficients, not the input.

If that sounds abstract, the intuition is simpler than the definition: think of R as the reach of the series. Inside that reach the series behaves itself and sums to a smooth function. At the edge, the series may or may not behave; you have to check each edge separately. Beyond the edge, the terms blow up and the series is meaningless.

The ratio test: the fastest way to find R

For most series you meet in practice, the radius of convergence comes from the ratio test. Take the limit of the ratio of consecutive term magnitudes:

R = lim_n→∞ |a_n / a_n+1|

The series converges when |x - c| < R and diverges when |x - c| > R. Why does this work? The ratio test compares the series to a geometric series term by term, and a power series is dominated by geometric behavior near the boundary of convergence.

Worked example: 1 / 2ⁿ

Take a_n = 1 / 2ⁿ, the series 1 + x/2 + x²/4 + .... This calculator evaluates the ratio at large n:

• n = 10: ratio = 2.0000
• n = 20: ratio = 2.0000
• n = 40: ratio = 2.0000

The ratio is exactly |a_n / a_n+1| = 2 for every n, because 1/2ⁿ ÷ 1/2ⁿ⁺¹ = 2. So R = 2, and the series converges for |x| < 2. This matches the fact that Σ (x/2)ⁿ is geometric in x/2.

Worked example: the harmonic series term 1/n

With a_n = 1/n, the ratio approaches 1 slowly, from above:

• n = 10: ratio = 1.1000
• n = 20: ratio = 1.0500
• n = 40: ratio = 1.0250

The pattern is (n + 1)/n = 1 + 1/n, which squeezes toward 1. So R = 1, and the interval is (-1, 1) with endpoint checks pending (see the endpoint section: this series converges conditionally at x = -1 and diverges at x = 1).

Worked example: n / 2ⁿ, where the ratio converges slowly

For a_n = n / 2ⁿ the ratio is |a_n/a_n+1| = 2n/(n+1):

• n = 10: ratio = 1.8182
• n = 20: ratio = 1.9048
• n = 40: ratio = 1.9512

The ratio creeps up toward 2 from below, so R = 2. This example shows why the calculator samples at n = 10, 20, 40, and beyond: the first step alone (1.82) would have led you to guess roughly 1.8. The limit is what matters.

Two degenerate cases of the ratio test

If the ratio limit is 0, then R = 0: the series converges nowhere except possibly at the center. The factorial term a_n = n! does this: the ratios are 0.0909, 0.0476, 0.0244, halving roughly each time, because n! / (n+1)! = 1/(n+1) collapses to zero. If the ratio grows without bound, then R = ∞: the series converges for every x. The term a_n = 1/n! gives ratios 11, 21, 41, growing linearly, so R = ∞, which is why the exponential series e^x = Σ xⁿ/n! works for all real x.

The root test: when the ratio test stalls

The root test is the other workhorse:

R = 1 / lim_n→∞ |a_n|^1/n

It gives the same radius whenever the ratio test works, and it keeps working in cases where the ratio is awkward or oscillates.

Worked example: 1 / 10ⁿ

Here |a_n|^1/n = 1/10 for every n, so the root test gives R = 1 / (1/10) = 10. The ratio test agrees: the ratios are 10.0000, 10.0000, 10.0000. The series Σ (x/10)ⁿ is geometric in x/10, so it converges for |x| < 10.

Worked example: 1 / nⁿ

This one is brutal for the ratio test and trivial for the root test. The root test evaluates |1/nⁿ|^1/n = 1/n, which crashes to zero, so R = ∞. The engine sees it directly:

• n = 10: root step = 10.0000
• n = 20: root step = 20.0000
• n = 40: root step = 40.0000

The root steps grow without bound, so the radius is infinite. The series Σ xⁿ/nⁿ converges for every real x.

Worked example: nⁿ, the opposite extreme

With a_n = nⁿ the root steps are 0.1000, 0.0500, 0.0250, halving toward zero, so R = 0. The series Σ nⁿxⁿ converges only at x = 0.

Which test should you use? Use the ratio test first when a_n contains factorials or products, because factorials telescope beautifully in ratios. Use the root test when a_n is a power like nⁿ or n^k. This calculator runs both and shows you both sets of steps, so you can see which one settles faster for your particular series.

Endpoints: turning the radius into the interval

The radius tells you the interval (c - R, c + R) where the series converges. The two endpoints x = c - R and x = c + R are separate questions, and this is where most students lose points: the ratio and root tests are inconclusive at the endpoints by construction, because the ratio there is exactly 1.

At an endpoint, the series becomes an ordinary numerical series. For example, the harmonic-based series with a_n = 1/n and c = 0 has R = 1, and the endpoints are x = -1 and x = 1:

• At x = 1 the series is 1 + 1/2 + 1/3 + ..., the harmonic series, which diverges.
• At x = -1 the series is 1 - 1/2 + 1/3 - 1/4 + ..., the alternating harmonic series, which converges, by the alternating series test, to the natural logarithm of 2.

So the interval of convergence is [−1, 1): closed at the left, open at the right. This calculator runs a partial-sum heuristic at both endpoints and returns likely-converges or likely-diverges for each, then assembles the bracket notation for you.

The three rules of endpoint behavior

1. Terms must tend to zero. If the terms of the series at the endpoint do not approach zero, the series must diverge there. The geometric series at x = 1 has terms 1, 1, 1, ...; the series Σ (1/2)ⁿ at x = 2 has terms 1, 1, 1, ...; and Σ n at x = 1 has terms 1, 2, 3, .... All three diverge, and the engine flags all three with the same message: terms do not tend to zero.
2. Alternating terms can save a borderline endpoint. If the terms shrink to zero monotonically and alternate in sign, the alternating series test guarantees convergence. That is exactly what happens at x = -1 for 1/n, 1/n^0.5, and 1/(n ln n): all three converge conditionally at the left endpoint even though the same series diverge at the right endpoint.
3. p-series are the reference standard. The series Σ 1/n^p converges when p > 1 and diverges when p ≤ 1. At the right endpoint of a centered-at-zero p-series, convergence and divergence hinge on this single comparison. The engine confirms the pattern: 1/n² converges at x = 1, while 1/n and 1/n^0.5 diverge there.

Absolute versus conditional convergence at endpoints

When a series converges at an endpoint, it may do so absolutely (the series of absolute values also converges) or conditionally (the series converges only because of the alternating signs). The distinction matters for rearrangement theorems, and it explains the shape of many intervals:

Notice the pattern: an alternating endpoint that is conditional is a left endpoint when the series is built from 1/n^p, because (-1)ⁿ appears at x = -1. Flip the sign of the coefficient, as in (-1)ⁿ/n, and the conditional endpoint flips to the right side.

Reference table: radius of convergence of common series

Every value below was produced by this page's engine, which evaluates the coefficients in logarithmic space so that factorials and nⁿ terms never overflow a double-precision number. Where the table lists an interval, both endpoints were checked numerically and the verdicts are shown.

Two things stand out in this table. First, most simple algebraic terms give R = 1: the radius only cares about the growth rate of the coefficients, and polynomials and logarithms grow too slowly to move the boundary away from 1. Second, exponential terms move the radius away from 1 in the direction you would guess: dividing by 2ⁿ doubles the radius, and multiplying by 2ⁿ halves it.

Series centered away from zero

Nothing changes except the location. The interval becomes (c - R, c + R) instead of (-R, R), and the endpoint values become x = c - R and x = c + R.

Take the p-series term a_n = 1/n² centered at c = 3. The radius is still 1, so the endpoints are x = 2 and x = 4. At x = 4 the series is Σ 1/n², which converges. At x = 2 the series is Σ (-1)ⁿ/n², which also converges (absolutely, since the absolute values are the same p = 2 series). The engine confirms both endpoints, and the interval is [2, 4].

Try it: load the preset 1/n^2 at c = 3 and watch the number line center on 3 instead of 0. The radius, the interval, and the endpoint dots all shift together.

When R = 0 or R = infinity

Both extremes are real, and both show up in the presets.

R = 0 means the series converges only at the center x = c (and even there only trivially). The factorial series Σ n! xⁿ is the classic example: no matter how small x is, the factorial growth eventually overwhelms xⁿ. The engine detects this when the ratio steps keep collapsing: 0.0909, 0.0476, 0.0244 for n!, or the root steps 0.1000, 0.0500, 0.0250 for nⁿ.

R = ∞ means the series converges for every real x, no matter how large. The exponential series Σ xⁿ/n! and the series Σ xⁿ/nⁿ both do this, for different reasons: factorial denominators shrink the terms faster than any power of x can grow, and nⁿ denominators do the same even more aggressively. In both cases the interval notation is (-∞, ∞) and there are no endpoints to check.

What numeric checking can and cannot prove

This calculator is honest about its limits, because the README of the project demands it. The ratio and root steps are exact where the ratio is constant (geometric-type series) and become exact estimates elsewhere. The endpoint verdicts come from a partial-sum heuristic: the engine sums terms in doubling blocks and checks whether the increment between blocks shrinks. That is strong evidence, but it is not a proof.

The clearest example of the limit is the series Σ 1/(n ln n) at its right endpoint x = 1. This series diverges, but it diverges slower than any computer can see: the partial sums grow like ln(ln N), and after two million terms the sum is only around 3.5. The doubling-block increments shrink so slowly that the numeric check returns inconclusive. The exact answer, by the integral test, is divergence. When you load that preset, the page tells you both things: what the numbers suggest, and what the test proves.

So treat the endpoint verdicts as a guided guess with a clear explanation of why, and use the classical tests (alternating series, p-series, comparison, integral) to lock in the final answer. The preset notes tell you the known answer for each built-in example so you can calibrate your trust in the numeric verdicts.

Common mistakes

• Forgetting that the ratio test uses |a_n/a_n+1|, not |a_n+1/a_n|. Flip the fraction and you get 1/R instead of R. The geometric series with ratio 2 would then look like it had radius 0.5. Every preset in this calculator is set up so you can catch that error yourself.
• Declaring the interval open at an endpoint without checking. The most common interval mistake in textbooks is writing (−1, 1) for the harmonic series, when the true answer is [−1, 1): the series converges conditionally at −1.
• Applying the ratio test at the endpoints. At x = c ± R the ratio is exactly 1, so the test is silent by design. The endpoints need the alternating series test, the p-series test, or the terms-not-zero rule.
• Forgetting that alternating signs appear at x = c - R. For a center of 0, the left endpoint introduces the factor (−1)ⁿ into every term. That factor is the difference between divergence and conditional convergence in half the examples in the table above.
• Confusing the radius with the interval. The radius is a single nonnegative number (or infinity). The interval is a set of x values. You cannot write "the interval of convergence is R = 1"; you write [−1, 1) or (0, 2) depending on the center.

Frequently asked questions

Key takeaways

• The radius of convergence R comes from the coefficients alone: R = lim |a_n/a_n+1| by the ratio test, or 1 / lim |a_n|^1/n by the root test. Both tests give the same R.
• The interval of convergence is (c - R, c + R) plus endpoints that pass their own tests. Always check x = c - R and x = c + R separately; the ratio test is inconclusive exactly there.
• Alternating signs at the left endpoint (for center 0) turn many divergent series into conditionally convergent ones: 1/n, 1/n^0.5, and 1/(n ln n) all converge at x = -1 and diverge at x = 1.
• R = 0 and R = ∞ are real outcomes, not errors: n! and nⁿ give R = 0; 1/n! and 1/nⁿ give R = ∞.
• Numeric endpoint checks are evidence, not proofs. For slow series like 1/(n ln n) at x = 1, use the integral test or comparison test to settle what the numbers cannot.

Conclusion

The radius of convergence is the first thing to compute for any power series, and the ratio and root tests give it to you quickly: divide consecutive terms or take n-th roots, read the limit, invert if needed. The radius alone does not finish the job. The interval of convergence is the radius plus a two-point investigation at x = c - R and x = c + R, where p-series comparisons, alternating series behavior, and the terms-not-zero rule decide whether each bracket is open or closed.

This radius of convergence calculator does the mechanical work and shows its work: ratio steps, root steps, endpoint verdicts, and a number line, all from the single expression a_n and the center c you type in. The numbers it produces match the classical results for every preset on the page, and it flags the one case in its preset list where the numeric evidence alone cannot decide. Type in your own series term, or start from a preset, and check whether your interval ends up open, closed, or mixed.