Power series — centre, radius of convergence, interval of convergence: Visual walkthrough

We will need three prerequisite ideas as we go — the Geometric Series (our model example), the Root Test and Ratio Test (our measuring tools), and Limit Superior and Limit Inferior (the tool that never fails). Each is introduced the moment it is used.

Step 1 — What a power series actually is

WHAT we just named: every ingredient of the sum. WHY it matters: because only $(x - a)^{n}$ depends on $x$ , the whole question "does it converge?" will come down to how big $(x - a)$ is — its size $∣ x - a ∣$ . PICTURE: think of the centre $a$ as a home base on the number line, and $x$ as a point some distance away.

At $x = a$ the distance is zero, so every $(x - a)^{n}$ with $n \geq 1$ vanishes and the sum is just $c_{0}$ . The centre is a free win — it always converges.

Step 2 — Build intuition on the simplest case: the geometric series

Before general coefficients, look at the cleanest possible power series, where every $c_{n} = 1$ and $a = 0$ : $\sum_{n = 0}^{\infty} x^{n} = 1 + x + x^{2} + x^{3} + \dots$ This is the Geometric Series. It is the seed from which the whole theory grows.

WHAT we did: computed a closed form and noted its domain. WHY this example: it already shows the punchline in miniature — convergence happens for $∣ x ∣ < 1$ (a symmetric interval around $0$ ) and fails outside. The magic number here is $R = 1$ . PICTURE: watch the terms $x^{n}$ for a shrinking $x$ versus a growing $x$ .

The left panel (terms collapse to zero) is convergence; the right panel (terms hold their size or grow) is divergence. Everything below is the general version of this single picture.

Step 3 — The right measuring tool: the Root Test

To handle any coefficients we need a test that reads a series and says "converge" or "diverge". We pick the Root Test.

WHAT the test measures: $n b_{n}$ is roughly the "common ratio felt at stage $n$ ". If that ratio is below $1$ the terms behave like a shrinking geometric series (Step 2), so the sum converges. WHY the cutoff is exactly $1$ : at $1$ the terms neither shrink nor grow decisively — precisely the borderline of Step 2's $∣ x ∣ < 1$ rule. PICTURE: $n b_{n}$ as a dial; below the line $1$ = safe, above = runaway.

Step 4 — Feed the power series into the Root Test

Now set $b_{n} = ∣ c_{n} (x - a)^{n} ∣ = ∣ c_{n} ∣ ∣ x - a ∣^{n}$ (we take absolute values because we want absolute convergence — see Absolute vs Conditional Convergence). Apply the root:

$n b_{n} = n coeff part ∣ c_{n} ∣ x part ∣ x - a ∣^{n} = depends only on n n ∣ c_{n} ∣ \cdot the n cancels! ∣ x - a ∣ .$

WHAT we did: separated the term into an $x$ -part and a coefficient-part. WHY it is the whole game: because $∣ x - a ∣$ pulls out, the limit is $L = ∣ x - a ∣ \cdot (lim_{n \to \infty} n ∣ c_{n} ∣) .$ The convergence condition $L < 1$ becomes a plain distance condition on $x$ . PICTURE: the term splits into two stacked bars, one fixed by $x$ , one fixed by the coefficients.

Step 5 — The coefficient number $β$ , and why we need $lim sup$

Call the coefficient limit $β = lim_{n \to \infty} n ∣ c_{n} ∣ .$ There is a snag: this ordinary limit may not exist (coefficients can oscillate). To repair it we use the Limit Superior and Limit Inferior.

So we replace $β$ by the always-defined $β = lim sup_{n \to \infty} n ∣ c_{n} ∣ .$

WHAT we did: swapped a possibly-missing limit for one that always exists. WHY: to make the formula work for every power series, even ones with jumpy coefficients (that is Example 5 in the parent note). PICTURE: an oscillating sequence — the ordinary limit is undefined, but the ceiling it keeps touching is crisp.

Step 6 — Solve the condition → the radius $R$

We now have $L = ∣ x - a ∣ \cdot β$ . Impose the Root Test's convergence rule $L < 1$ :

$∣ x - a ∣ \cdot β < 1 ⟺ ∣ x - a ∣ < \frac{1}{β} .$

Name that boundary distance $R$ : $\frac{1}{R} = n \to \infty lim sup n ∣ c_{n} ∣ (Cauchy-Hadamard) .$

$\frac{1}{R} = β$ , with conventions $1/0 = \infty$ and $1/\infty = 0$ .

PICTURE: a coloured band of half-width $R$ around $a$ ; inside = converge, outside = diverge.

Step 7 — The three regimes of $R$ (cover every case)

$R$ can be any value in $[0, \infty]$ . All three behave differently:

| $R$ | $β = lim sup n ∣ c_{n} ∣$ | Where it converges | Example | |---|---|---|---| | $0$ | $\infty$ | only at $x = a$ | $\sum n! x^{n}$ | | $0 < R < \infty$ | finite $> 0$ | band $(a - R, a + R)$ + maybe ends | $\sum \frac{( x - 2 ) ^{n}}{n}$ | | $\infty$ | $0$ | all real $x$ | $\sum \frac{x ^{n}}{n !} = e^{x}$ |

WHAT / WHY: the coefficient ceiling $β$ inversely controls reach. Fast-growing coefficients ( $β = \infty$ ) choke the series to a single point; fast-shrinking coefficients ( $β = 0$ ) let it converge everywhere. PICTURE: three number lines — a single dot, a finite band, the whole line.

Step 8 — The endpoints: the one place the machine goes quiet

At exactly $∣ x - a ∣ = R$ we get $L = R \cdot β = 1$ . That is the Root/Ratio Test's inconclusive value. The machine shrugs; we must test by hand.

At $x = a + R$ and $x = a - R$ the series becomes a plain number series (no more $x$ ). Each endpoint independently converges or diverges, so the interval is one of $(a - R, a + R), [a - R, a + R), (a - R, a + R], [a - R, a + R] .$

PICTURE: the band with two hollow/solid dots to be decided one at a time.

The one-picture summary

Everything above in a single flow: split the term, take the root, cancel the power, read off the band, hand-check the ends.

Recall Feynman retelling — say it in plain words

Imagine standing at the centre $a$ on a number line. A power series asks: "how far can I walk before the infinite sum falls apart?" The only $x$ -flavoured piece of each term is $(x - a)^{n}$ , a distance raised to a power. To measure such a thing you take the $n$ -th root — and the root eats the power, leaving just the raw distance $∣ x - a ∣$ times a pure coefficient number $β$ (the ceiling of the coefficient roots, which we grab with $lim sup$ so it always exists). The series survives while distance $\times β$ stays under $1$ , i.e. while $∣ x - a ∣ < 1/ β = R$ . A distance-under- $R$ rule always paints a symmetric band, so the answer is always an interval. Big coefficients shrink the band to a dot ( $R = 0$ ); tiny ones open it to the whole line ( $R = \infty$ ). Right at the edge the tests go silent, so we walk to each endpoint, plug the number in, and test that lone series on its own.

Recall Self-check

Why does the $x$ -dependence separate out cleanly in the root test? ::: Because $n ∣ x - a ∣^{n} = ∣ x - a ∣$ exactly — root and power cancel, leaving a single factor. Why $lim sup$ instead of $lim$ ? ::: The ordinary limit of $n ∣ c_{n} ∣$ may not exist, but $lim sup$ always exists in $[0, \infty]$ , so the formula is universal. Why is the convergence set always an interval? ::: The condition is $∣ x - a ∣ < R$ , a distance-from-centre bound, whose solution is the symmetric band $(a - R, a + R)$ . Why must endpoints be tested separately? ::: At $∣ x - a ∣ = R$ the test limit equals $1$ , the inconclusive case, so each end needs its own number-series test.

Related: Taylor and Maclaurin Series · Term-by-term Differentiation and Integration · 4.3.14 Power series — centre, radius of convergence, interval of convergence (Hinglish)