What the radius of convergence tells you
A power series Σ aₙ(x − c)ⁿ is a function only where it converges, and the set on which it converges is always an interval centred at c. That is the substantial theorem here: convergence is never patchy. There is a number R, possibly 0 and possibly unbounded, such that the series converges absolutely for |x − c| < R, diverges for |x − c| > R, and does something that has to be checked case by case at the two endpoints.
That number matters because it bounds where a Taylor series is usable. The geometric series Σxⁿ equals 1/(1 − x) only on (−1, 1); feeding it x = 2 gives the notorious "1 + 2 + 4 + … = −1". The Maclaurin series for ln(1 + x) converges on (−1, 1], so it computes ln 2 at the right endpoint and nothing at all at x = 2. Any numerical method that truncates a series inherits its radius as a hard limit.
The radius also carries information about the function itself. For a real analytic function the radius equals the distance from the centre to the nearest singularity in the complex plane. That is why 1/(1 + x²), which is perfectly smooth for every real x, still has a Maclaurin series that stops working at |x| = 1: the poles sit at ±i, one unit away. The real graph gives no warning at all.
The ratio test, the root test, and where each one wins
Apply the ratio test to the terms of the series, not to the coefficients alone. The n-th term is aₙ(x − c)^(pn), so the ratio of consecutive magnitudes is |aₙ₊₁/aₙ|·|x − c|^p. Convergence requires that to be below 1, which rearranges to |x − c|^p < lim|aₙ/aₙ₊₁|. With p = 1 the radius is the ratio limit itself; with p = 2 it is the square root of it.
The root test gives the same radius by a different route: R = 1/limsup|aₙ|^(1/n), which is the Cauchy–Hadamard theorem. It is the stronger statement, because a limsup always exists while the ratio limit may not — a series whose coefficients alternate between two patterns can defeat the ratio test entirely. In exchange it converges very slowly numerically, since n^(1/n) creeps towards 1 like 1 + (ln n)/n. This page reports both and treats the ratio value as primary, which is the right choice for coefficients built from powers, polynomials and factorials.
Neither test says anything at the endpoints, because the ratio is exactly 1 there and the test is inconclusive by construction. Endpoints have to be handled by a test that sees the terms themselves. This calculator estimates the exponent q in |aₙRⁿ| ≈ n^(−q) from the slope of the terms on a log-log scale between two widely separated indices, then applies the standard criteria: q > 1 means absolute convergence by comparison with a p-series; q ≤ 1 with terms that alternate in sign and shrink monotonically means conditional convergence by the alternating series test; anything else diverges.
That is a numerical estimate, not a proof, and the page says so. It is reliable for the coefficient families that appear in coursework and it will mislead you on a coefficient with a logarithmic factor, where the true behaviour is 1/(n ln n) and the fitted exponent reads slightly above 1 while the series still diverges. Confirm an endpoint verdict by hand before you quote it.
Worked example: Σ (x − 3)ⁿ / n²
Take aₙ = 1/n², centre c = 3, ordinary powers so p = 1.
- Form the ratio. |aₙ/aₙ₊₁| = (1/n²)·(n+1)² = (n+1)²/n² = (1 + 1/n)².
- Take the limit. As n → ∞, (1 + 1/n)² → 1. So R = 1.
- Write the open interval. |x − 3| < 1, that is 2 < x < 4.
- Test the right endpoint, x = 4. The series becomes Σ 1ⁿ/n² = Σ 1/n², a p-series with p = 2 > 1. It converges — to π²/6, in fact. Include the endpoint.
- Test the left endpoint, x = 2. The series becomes Σ(−1)ⁿ/n², which converges absolutely because Σ|(−1)ⁿ/n²| = Σ1/n² converges. Include it too.
- State the interval of convergence. [2, 4], closed at both ends.
Now change one thing: use aₙ = 1/n instead, still centred at 3. The ratio is (n+1)/n → 1, so R = 1 again and the open interval is unchanged. But the right endpoint now gives the harmonic series Σ1/n, which diverges, while the left endpoint gives Σ(−1)ⁿ/n, which converges conditionally to −ln 2. The interval is [2, 4) — half open. Two series with the same radius and different intervals, which is exactly why endpoints get their own step.
Check the decay exponent the calculator reports against this. For aₙ = 1/n² at the endpoint, the terms are 1/n², so q = 2; the log-log slope between n = 40 and n = 160 is [ln(1/1600) − ln(1/25600)]/ln 4 = ln(16)/ln(4) = 2 exactly. For aₙ = 1/n the same calculation gives ln(4)/ln(4) = 1. The threshold q > 1 separates them correctly.
Reading the radius, the interval and the endpoints
Read the radius first. A finite positive radius is the ordinary case and gives you an interval with two endpoints to check. A radius reported as unbounded means the ratio grows without limit — the signature of a factorial in the denominator — and the series converges for every real x, as the exponential, sine and cosine series do. A radius of zero means the ratio collapses to zero, the signature of a factorial in the numerator, and the series is useless except at the centre itself.
Then read the two endpoint verdicts together, because they are often different and the difference is the whole point of the exercise. A series can converge absolutely at both ends, diverge at both, or converge at one and diverge at the other. What it cannot do is converge conditionally at both ends with p = 1, because the terms at one end are the terms at the other with alternating signs, and alternating signs are what makes conditional convergence possible.
The decay exponent is the number doing the work. Values comfortably above 1 mean absolute convergence at both endpoints and a closed interval. Values at or below 0 mean the terms do not even tend to zero, so both endpoints diverge by the divergence test — that is the geometric case, where aₙRⁿ is constant. Values between 0 and 1 are the interesting band: divergence on the non-alternating side, conditional convergence on the alternating side.
Finally, compare the two radii. Large disagreement between the ratio and root estimates is normal and not an error: the root test estimate at n = 160 for aₙ = 1/n is about 1.032 rather than 1, because n^(1/n) converges too slowly to do better. Treat the root value as a cross-check on the order of magnitude, and the ratio value as the answer.
Reference: standard series, their radii and their intervals
| Series | aₙ | R | Interval of convergence |
|---|---|---|---|
| Σ xⁿ (geometric) | 1 | 1 | (−1, 1) |
| Σ xⁿ/n | 1/n | 1 | [−1, 1) |
| Σ xⁿ/n² | 1/n² | 1 | [−1, 1] |
| Σ xⁿ/n! (eˣ) | 1/n! | unbounded | (−∞, ∞) |
| Σ n! xⁿ | n! | 0 | {0} only |
| Σ xⁿ/2ⁿ | 1/2ⁿ | 2 | (−2, 2) |
| Σ n²xⁿ/3ⁿ | n²/3ⁿ | 3 | (−3, 3) |
| Σ (−1)ⁿx²ⁿ/(2n)! (cos x) | (−1)ⁿ/(2n)! | unbounded | (−∞, ∞) |
| Σ (x−3)ⁿ/n² | 1/n² | 1 | [2, 4] |
The three rows with R = 1 have the same radius and three different intervals — open, half-open and closed. That is why the endpoints are checked separately and why the radius alone is never a complete answer.
Pitfalls and the limits of the numerical approach
- Reporting the radius as the answer. The question usually asks for the interval of convergence, which requires both endpoints tested. A radius with no endpoint work is an incomplete answer.
- Applying the ratio test to aₙ instead of to the full term. The (x − c)^(pn) factor is what makes the inequality depend on x at all. Leaving it out loses the p-th root when p ≠ 1.
- Assuming a series in even powers has radius equal to the ratio limit. With p = 2 the ratio limit bounds |x − c|², so the radius is its square root.
- Trusting the numerical endpoint verdict on a borderline coefficient. A factor of ln n shifts the fitted exponent slightly without changing the divergence, and no finite sample can see the difference. Verify borderline cases by hand.
- Overflow in factorials. Double precision stops representing n! above n = 170. The calculator halves its sample index automatically when that happens, but very large sample settings will still fail.
- Expecting a coefficient with two interleaved patterns to work. If aₙ is zero for odd n, the ratio is undefined at every other index. Rewrite the series so that every coefficient is non-zero and use the exponent multiplier.
Endpoints are where absolute and conditional convergence separate
Inside the interval, convergence is always absolute. Exactly at an endpoint it may be merely conditional, which means the series converges but the series of absolute values does not — and a conditionally convergent series can be rearranged to sum to any value you like, by Riemann's rearrangement theorem. That is not a curiosity: it means term-by-term manipulation that is safe inside the interval is unsafe at an endpoint. If you need to resolve an indeterminate limit built from such a series, the L'Hôpital's rule calculator works from derivatives instead and sidesteps the issue.
Where the radius comes from and what it constrains
The radius is the bridge between series and functions. Term-by-term differentiation and integration of a power series are both valid on the open interval and both preserve the radius exactly — the derivative of Σaₙxⁿ is Σn aₙxⁿ⁻¹ with the same R, even though the coefficients have changed. What can change is endpoint behaviour: differentiating Σxⁿ/n² gives Σxⁿ⁻¹/n, and the closed interval becomes half-open.
In practice the radius sets the useful range of a truncated series. A Taylor polynomial used for numerical evaluation is trustworthy well inside the interval and useless outside it, and near the boundary the number of terms needed grows without limit. That is the same trade-off you meet when choosing a step size for a numerical derivative or a panel count for a Riemann sum: an error that shrinks with effort until something else stops it.
Series also give the cleanest route to many limits. Expanding numerator and denominator and comparing leading coefficients resolves a 0/0 form in one step, which is often faster than repeated differentiation — the two approaches are compared on the L'Hôpital's rule page. And knowing where a series converges tells you where that substitution is legitimate in the first place.
One assumption is worth naming: this calculator treats x as real. The theory is genuinely complex-analytic — the region of convergence is a disc in the complex plane and the radius is its radius, which is where the word comes from. On the real line you see only the diameter of that disc, and the singularity that determines it may be nowhere near the real axis.