4.3.14 · D5Calculus III — Sequences & Series
Question bank — Power series — centre, radius of convergence, interval of convergence
The tools you will be reasoning about live here: Ratio Test, Root Test, Limit Superior and Limit Inferior, Geometric Series, p-series, Alternating Series Test, Absolute vs Conditional Convergence, Taylor and Maclaurin Series, Term-by-term Differentiation and Integration, and the parent Power series topic note.
Vocabulary refresher (so every trap makes sense)
Picture the number line below: the centre sits in the middle, is the reach in each direction, and the two dots are the endpoints that must always be checked by hand.
True or false — justify
A power series always converges at its own centre.
True. Setting kills every term except (all others have a factor ), so the sum is just the single number — convergence is automatic there.
If the series diverges everywhere.
False. means it converges only at the centre (where it gives ); it diverges at every other point, but the centre itself still works.
The interval of convergence is always symmetric about the centre.
The open part is symmetric — — because is a distance condition. But the closed interval need not be symmetric in behaviour: one endpoint can converge while the other diverges (as in ).
If a power series converges at and its centre is , it must converge at .
True. Convergence at means it converges at distance from the centre, so ; the point is at distance , safely inside, hence it converges there (absolutely, in fact).
If a power series converges at (centre ), it must converge at .
Not guaranteed. could be the endpoint (distance exactly ). Then lies outside the radius, where the series diverges. Convergence at a point only guarantees convergence strictly closer to the centre.
The radius of convergence can be any number in .
True. (converges only at centre), (converges everywhere, like the series), and every finite positive value all occur.
A power series and its term-by-term derivative have different radii of convergence.
False. Term-by-term Differentiation and Integration preserves — differentiating or integrating a power series gives one with the same radius (only the endpoint behaviour may change).
The Ratio Test formula works for every power series.
False. It only works when that limit exists. If the coefficient ratio oscillates without settling (see the parent's Example 5), the formula gives nothing and you must use the always-valid root formula.
always exists (possibly ).
True. That is the whole point of using Limit Superior and Limit Inferior instead of an ordinary limit: the is defined for every bounded-below sequence, taking a value in , so Cauchy–Hadamard never breaks.
Spot the error
"The ratio limit at the endpoints equalled , and the Ratio Test says converges, so the endpoints converge."
Wrong. A limit of exactly is the inconclusive case of the Ratio Test, not the "less than " case. You learn nothing about the endpoints from it — substitute each one and run a separate test.
" has because grows so fast."
Backwards. Fast-growing coefficients make the series harder to converge, not easier. The ratio for any , so — it converges only at .
"The centre of is ."
Sign error. The form is , so means the centre is ====. The centre is the value of that makes the base zero.
"I found and the open interval is , so the answer is ."
Incomplete. You cannot state the interval until you test both endpoints and by substitution; the true interval may include one, both, or neither.
" converges, so converges absolutely."
Conflates two ideas. It converges conditionally by the Alternating Series Test, but (the absolute version) diverges. See Absolute vs Conditional Convergence — inside convergence is absolute, but at an endpoint it may only be conditional.
"At an endpoint the series always either converges at both or diverges at both."
False claim of symmetry. Endpoints are two different number series and behave independently — diverges at (harmonic) but converges at (alternating).
"Since equals , and is finite for all , the series must converge everywhere — that's why ."
The conclusion is right but the reasoning is circular. We establish first by the Ratio Test (); that is what licenses calling the sum for all . You cannot assume the identity to prove the convergence that justifies the identity.
Why questions
Why is the convergence set always an interval and never, say, two separate pieces?
Because the convergence condition is — a single distance-from-a-point inequality. Its solution set is one symmetric band around , which is exactly what "interval" means; a distance condition cannot carve out disconnected chunks.
Why do we use rather than in the Cauchy–Hadamard formula?
Because may fail to exist (oscillating coefficients), whereas always exists in . Using it makes the formula valid for every power series, not just the well-behaved ones.
Why does the factor pull cleanly out of the Root Test?
Because exactly for every — it is a fixed number independent of . So the limit only has to deal with the coefficient part , and rides along untouched.
Why must endpoints be tested separately instead of being handled by the same test that gave ?
Both the Ratio and Root Tests give a limit of exactly at , their one blind spot. So a fresh, sharper tool (p-series, Alternating Series Test, divergence test) is needed to resolve each endpoint.
Why does the Ratio version give — reciprocal of the coefficient ratio?
Convergence needs , i.e. . The boundary of that inequality is , so comes out as the reciprocal.
Why can a Geometric Series like tell us its radius at a glance?
Its coefficients are all , so and ; equivalently it is the model case that converges iff and diverges at both endpoints . It is the prototype every power-series radius is measured against.
Edge cases
If every coefficient past some point, what is the series and its radius?
It becomes an ordinary polynomial (finitely many terms). Apply the Root Test: past the cutoff , so and , giving — it converges for all .
What is the interval of convergence when ?
The whole real line — there are no endpoints to test, since is never attained.
If a power series converges only at its centre, what is its interval of convergence?
The single point , corresponding to . It is a degenerate "interval" of zero length.
Can both endpoints diverge while the open interval converges?
Yes. E.g. has and both endpoints give , which diverge (terms don't ), so the interval is the open .
Can both endpoints converge, giving a fully closed interval?
At an endpoint, can the series converge conditionally but not absolutely?
Yes. At , becomes : conditionally convergent (alternating) but not absolutely (harmonic). Inside the radius convergence is always absolute; only at endpoints can it drop to conditional.
What happens to the endpoints when we differentiate a power series term-by-term?
The radius is unchanged, but an endpoint that previously converged may start to diverge (differentiation divides in an extra factor of , degrading convergence). The interval can only shrink at the ends, never grow. See Term-by-term Differentiation and Integration.
If the centre is complex or the coefficients are strange, is still well-defined?
Yes. Because uses only the magnitudes and a that always exists, a radius is defined for any coefficient sequence whatsoever.
Recall One-line summary of the traps
Trap ::: The three habits that bite: forgetting to test endpoints, trusting the Ratio limit when it doesn't exist, and confusing convergence-at-a-point with convergence-everywhere-closer. Reason from every time.