Power series, radius of convergence, Abel's theorem

In one sentence

A power series has a radius inside which everything works and outside which nothing does, and the whole subject is what happens on the circle in between.

Why the exam asks it

Because the radius is computed with a limsup, not a limit — so a candidate who reaches for the ratio test will be right most of the time and wrong exactly when the paper wants them to be. And because boundary behaviour is decided separately at every point, which gives an examiner four independent options from one series.

The idea

For $\sum a_n x^n$, the Cauchy–Hadamard formula gives the radius of convergence:

$$\frac{1}{R} = \limsup_{n \to \infty} |a_n|^{1/n}.$$

Inside $|x| < R$ the series converges absolutely. Outside $|x| > R$ it diverges. At $|x| = R$ anything can happen, and it must be checked point by point.

Note that this is a limsup, so it always exists in $[0, \infty]$ — the formula never fails. The ratio version, $R = \lim |a_n / a_{n+1}|$, is easier to compute and requires that limit to exist, which it often does not.

What survives inside the radius

On any closed sub-interval $[-r, r]$ with $r < R$, convergence is uniform. That is what lets you differentiate and integrate term by term, and both operations leave the radius unchanged:

$$\sum a_n x^n, \qquad \sum n a_n x^{n-1}, \qquad \sum \frac{a_n}{n+1}x^{n+1}$$

all have the same $R$. The behaviour at the endpoints can change, though, and that is examined directly.

The boundary

$\sum x^n$, $\sum x^n/n$ and $\sum x^n/n^2$ all have $R = 1$ and behave completely differently on $|x| = 1$: the first diverges everywhere on the circle, the second converges at $x = -1$ and diverges at $x = 1$, the third converges everywhere. Same radius, three different answers.

Abel's theorem is the one result that reaches the boundary: if $\sum a_n$ converges, then $\sum a_n x^n \to \sum a_n$ as $x \to 1^-$. It runs in that direction only.

Where intuition breaks

"The radius is $\lim |a_n/a_{n+1}|$." Only when that limit exists. Take a series whose coefficients alternate between two patterns — the ratios oscillate and have no limit, while $\limsup |a_n|^{1/n}$ is perfectly well defined. A limit assumed to exist, and the reason Cauchy–Hadamard is stated with a limsup at all.

"$\sum a_n x^n \to L$ as $x \to 1^-$, so $\sum a_n = L$." False without a further condition. Take $a_n = (-1)^n$: the series sums to $1/(1+x)$ inside the disc, which tends to $1/2$ as $x \to 1^-$. But $\sum (-1)^n$ diverges — it has no sum at all.

Why it is tempting: Abel's theorem looks like a statement about continuity at the boundary, and continuity feels reversible. It is not. The converse needs a Tauberian hypothesis such as $n a_n \to 0$; without one, the limit can exist while the series does not converge.

"A power series converges uniformly on its open disc of convergence." False, and this catches good candidates. $\sum x^n$ on $(-1,1)$: the tail is $x^{N+1}/(1-x)$, whose supremum over $(-1,1)$ is infinite for every $N$. Uniform convergence holds on every closed $[-r,r]$ with $r < 1$, and not on the open interval itself.

The distinction is the same pointwise versus uniform failure as elsewhere: uniformity on each compact piece does not glue into uniformity on the union.

"Differentiating preserves the behaviour at the endpoints." It preserves the radius, not the endpoint behaviour. $\sum x^n/n^2$ converges at $x = 1$; its derivative series $\sum x^{n-1}/n$ diverges there. Differentiation costs you one power of $n$, and at the boundary that is exactly the margin.

"$R = 0$ means the series is useless." It means it converges only at $x = 0$ — which is still a fact worth stating, and appears as a correct option often enough that rejecting it on instinct is a mistake. $\sum n! x^n$ is the standard example.

The exam's angle

1. Compute $R$ with limsup, and check whether the ratio limit exists before using the easier formula.
2. Test each endpoint separately. They are two independent questions and the paper often makes them differ.
3. If the option says "uniformly on $(-R, R)$", suspect it. The true statement is "on every closed subinterval".
4. If Abel appears, check the direction. Assuming the series converges is the hypothesis; concluding it is the error.

The night before

• $1/R = \limsup |a_n|^{1/n}$. Always exists; the ratio formula may not.
• Inside: absolute convergence. Outside: divergence. On the circle: check each point.
• Uniform convergence on every $[-r,r]$ with $r < R$, not on $(-R,R)$. Spoiler: $\sum x^n$.
• Term-by-term differentiation and integration preserve $R$, not endpoint behaviour.
• Abel: $\sum a_n$ convergent $\Rightarrow$ the limit from inside equals it. No converse — $\sum (-1)^n x^n \to 1/2$ while $\sum(-1)^n$ diverges.
• $\sum n! x^n$ has $R = 0$; $\sum x^n/n!$ has $R = \infty$.