Term-by-term differentiation and integration of power series: Visual walkthrough

Step 0 — What is a power series, in pictures?

WHAT. A power series is an endless sum $f (x) = \sum_{n = 0}^{\infty} a_{n} (x - c)^{n} = a_{0} + a_{1} (x - c) + a_{2} (x - c)^{2} + \dots$

Let me name every piece, right where it sits:

$a_{n}$ ::: the coefficient — a fixed number for each step $n$ (the "weight" of that term).
$c$ ::: the centre — the point we build around.
$(x - c)^{n}$ ::: the distance from the centre, raised to a power — it grows fast when $x$ is far from $c$ .
$\sum_{n = 0}^{\infty}$ ::: "add these up forever, starting at $n = 0$ ".

WHY care about a "radius". If $x$ is close to $c$ , the powers $(x - c)^{n}$ shrink and the sum settles on a number. If $x$ is far, the powers explode and the sum blows up. The border between "settles" and "explodes" is a distance $R$ from the centre, called the radius of convergence. Inside that distance the series is a real function; outside it is garbage.

PICTURE. A number line with the centre $c$ in the middle and a red "safe zone" of half-width $R$ on either side.

Figure — Term-by-term differentiation and integration of power series

Step 1 — What does differentiating do to each term?

WHAT. Apply $\frac{d}{d x}$ (the slope-finder) to one term $a_{n} (x - c)^{n}$ . The power rule says: bring the power down in front, then lower the power by one. $\frac{d}{d x} [a_{n} (x - c)^{n}] = n a_{n} (x - c)^{n - 1}$

Term by term, what changed:

the factor $n$ ::: new, pulled down from the exponent — this is the whole story.
$(x - c)^{n - 1}$ ::: the power dropped by one.
the $n = 0$ term ( $a_{0}$ , a constant) ::: differentiates to $0$ and disappears, so the sum now starts at $n = 1$ .

WHY this tool. We want the slope of $f$ . For an ordinary short polynomial the power rule is the machine that gives slopes. The claim of this chapter is that the endless polynomial obeys the same rule — so the differentiated series is $f^{'} (x) = \sum_{n = 1}^{\infty} n a_{n} (x - c)^{n - 1} .$

PICTURE. A tower of coefficients; differentiation multiplies the $n$ -th brick by $n$ and shifts the tower down one step.

Step 2 — What does integrating do to each term?

WHAT. Apply $\int \cdot d x$ (the area-finder) to one term. The rule: raise the power by one, then divide by the new power. $\int a_{n} (x - c)^{n} d x = \frac{a _{n}}{n + 1} (x - c)^{n + 1} + C$

Term by term:

the divisor $n + 1$ ::: new — integrating makes coefficients smaller.
$(x - c)^{n + 1}$ ::: the power rose by one.
$C$ ::: one single constant for the whole sum (the height the antiderivative starts at).

So the integrated series is $\int f (x) d x = C + \sum_{n = 0}^{\infty} \frac{a _{n}}{n + 1} (x - c)^{n + 1} .$

WHY. Same reason: for a short polynomial this is the rule for antiderivatives, and we claim the endless one plays along.

PICTURE. The same coefficient tower, but now the $n$ -th brick is divided by $n + 1$ and the tower shifts up one step — the mirror image of Step 1.

Step 3 — The machine that measures $R$

WHAT. The Cauchy–Hadamard theorem gives the radius directly from the coefficients: $\frac{1}{R} = lim sup_{n \to \infty} ∣ a_{n} ∣^{1/ n} .$

Decoding every symbol:

$∣ a_{n} ∣^{1/ n}$ ::: the $n$ -th root of the size of the $n$ -th coefficient. It asks: "roughly, what constant base $b$ makes $∣ a_{n} ∣ \approx b^{n}$ ?"
$lim sup_{n \to \infty}$ ::: the "eventual ceiling" of those roots as $n$ grows — the largest value they keep coming back near.
$\frac{1}{R}$ ::: one over the radius. Big root ⇒ big $1/ R$ ⇒ small safe zone. Small root ⇒ large safe zone.

WHY this tool and not another. The Geometric series converges exactly when its base has size below $1$ . Cauchy–Hadamard says a power series behaves like a geometric series whose base is $∣ a_{n} ∣^{1/ n} ∣ x - c ∣$ . So the radius is set by the root behaviour of the coefficients — nothing else. This is precisely the tool that turns "does it settle?" into a single number.

PICTURE. The roots $∣ a_{n} ∣^{1/ n}$ plotted as dots; a horizontal red line marks the ceiling they hover under. That ceiling is $1/ R$ .

Step 4 — The one limit that runs the whole engine

WHAT. We will need this single fact: $lim_{n \to \infty} n^{1/ n} = 1.$ In words: the $n$ -th root of $n$ marches to $1$ .

WHY we need it. Differentiation slipped an extra factor $n$ into each coefficient. Cauchy–Hadamard cares about $∣ a_{n} ∣^{1/ n}$ . So the extra factor contributes an extra $n^{1/ n}$ to the root. If that extra factor tends to $1$ , it multiplies the ceiling by $1$ — i.e. it changes nothing. This limit is the whole reason the radius survives.

WHY it's true (take logs). A quantity of the shape " $\infty^{0}$ " — here a growing $n$ raised to the shrinking power $1/ n$ — is a tug-of-war. Logarithms turn the power into a product where we can see who wins: $ln (n^{1/ n}) = \frac{1}{n} ln n = \frac{l n n}{n} .$

$ln n$ ::: grows, but slowly (logarithm crawls).
$n$ ::: grows in a straight line. As $n \to \infty$ , the straight-line denominator crushes the crawling numerator, so $\frac{ln n}{n} \to 0$ . Undo the log: $n^{1/ n} = e^{l n n / n} \to e^{0} = 1.$

PICTURE. Two racing curves: $ln n$ (crawling) versus $n$ (sprinting). Their ratio, in red, sinks to $0$ — dragging $n^{1/ n}$ down to $1$ .

Step 5 — Put it together: the differentiated radius equals $R$

WHAT. The differentiated series has coefficients $b_{n} = (n + 1) a_{n + 1}$ (re-index so both series start at $n = 0$ ). Its radius $R^{'}$ obeys $\frac{1}{R ^{'}} = lim sup_{n \to \infty} (n + 1) a_{n + 1}^{1/ n} .$

Split the root into two factors — this is the payoff: $(n + 1) a_{n + 1}^{1/ n} = \to 1 (Step 4) (n + 1)^{1/ n} \cdot same ceiling as ∣ a_{n} ∣^{1/ n} ∣ a_{n + 1} ∣^{1/ n} .$

$(n + 1)^{1/ n} \to 1$ ::: same reason as $n^{1/ n}$ ; shifting by one inside a root that goes to $1$ still goes to $1$ .
$∣ a_{n + 1} ∣^{1/ n}$ ::: the coefficients' own root, but a subtlety hides here — read the two callouts below.

Multiply the differentiation factors: $\frac{1}{R ^{'}} = 1 \cdot \frac{1}{R} = \frac{1}{R}$ , so $R^{'} = R$ . The integration case (Subtlety B) lands on the same $R$ by the reciprocal factor. The safe zone is untouched either way. ∎

PICTURE. The coefficient-root ceiling from Step 3, with the extra factor drawn as a red multiplier that shrinks to $1$ , leaving the ceiling — and hence $R$ — exactly where it was.

Step 5½ — WHY are we even allowed to swap $\sum$ and $\frac{d}{d x}$ ?

WHAT. Everything so far assumed we may differentiate the endless sum one term at a time: $\frac{d}{d x} \sum_{n} a_{n} (x - c)^{n} = ? \sum_{n} \frac{d}{d x} [a_{n} (x - c)^{n}] .$ This swap is not free for a general infinite sum of functions — you can build sums where it fails. So we must earn it for power series.

WHY it holds here. The licence is uniform convergence. The theorem from analysis is: if a series of differentiable functions converges at one point, and the series of their derivatives converges uniformly, then you may differentiate term by term. "Uniformly" means the tail of the sum can be made small by one bound that works for every $x$ at once — the pieces settle in lock-step, not at different speeds for different $x$ .

Why power series pass the test. Fix any closed sub-interval $∣ x - c ∣ \leq ρ$ with $ρ < R$ — strictly inside the safe zone. On it, each derivative term is bounded: $n a_{n} (x - c)^{n - 1} \leq = M_{n} (no x in it!) n ∣ a_{n} ∣ ρ^{n - 1} .$

$M_{n}$ ::: a pure number for each $n$ — the biggest that term can be anywhere on the sub-interval.

Why $\sum M_{n} < \infty$ (comparison to a geometric series). Pick a number $σ$ strictly between $ρ$ and $R$ , so $ρ < σ < R$ . Because $σ < R$ , the differentiated series' root ceiling (Step 5) gives $∣ a_{n} ∣^{1/ n} < \frac{1}{σ}$ for all large $n$ , i.e. $∣ a_{n} ∣ σ^{n} < 1$ , so $∣ a_{n} ∣ < σ^{- n}$ . Substitute: $M_{n} = n ∣ a_{n} ∣ ρ^{n - 1} < \frac{n}{ρ} (\frac{ρ}{σ})^{n} .$ Now $r := \frac{ρ}{σ} < 1$ , so $\sum_{n} n r^{n}$ is a convergent series (a geometric series $\sum r^{n}$ nudged by the polynomial factor $n$ — and $\sum n r^{n} = \frac{r}{( 1 - r ) ^{2}} < \infty$ for $∣ r ∣ < 1$ ; the factor $n$ is crushed by the geometric decay). Hence $\sum M_{n} < \infty$ : a finite bound.

The Weierstrass $M$-test then says: a single finite bound $\sum M_{n} < \infty$ forces the derivative series to converge uniformly on $∣ x - c ∣ \leq ρ$ . That is exactly the hypothesis the swap-theorem needed. Since every point strictly inside $(c - R, c + R)$ lies in some such $ρ$ -interval, the swap is legal everywhere inside the safe zone.

The integration swap is easier: uniform convergence of the original series on $∣ x - c ∣ \leq ρ$ (same $M$ -test, now with $M_{n} = ∣ a_{n} ∣ ρ^{n} < r^{n}$ , a plain convergent geometric series) already licenses swapping $\int$ and $\sum$ . So Step 5's radius result is not a bonus — it is the very thing that feeds the $M$ -test and makes term-by-term calculus honest.

Step 6 — The edge cases: endpoints, and the extremes $R = 0$ , $R = \infty$

WHAT. Everything above is about the open safe zone $∣ x - c ∣ < R$ . Three edge situations remain: the two boundary points, and the two extreme radii.

Endpoints can move. The points $x = c - R$ and $x = c + R$ are a separate story. There the geometric protection is gone (base $= 1$ ), so a leftover factor of $n$ or $\frac{1}{n + 1}$ suddenly matters.

$n = 1 \sum \infty n x^{n - 1}$ at $x = 1$ is $1 + 2 + 3 + \dots$ ::: diverges (differentiation lost the endpoint).
The $arctan$ build $\sum \frac{( - 1 ) ^{n}}{2 n + 1}$ at $x = 1$ converges (alternating) ::: integration gained an endpoint.

This is Abel's theorem (endpoint behaviour) territory, and why the parent's mnemonic DIRT ends in "Test endpoints again."

Extreme $R = 0$ survives. If the original series converges only at $x = c$ , then $lim sup ∣ a_{n} ∣^{1/ n} = \infty$ . The differentiated ceiling is $lim sup (n + 1)^{1/ n} ∣ a_{n + 1} ∣^{1/ n} = 1 \cdot \infty = \infty$ , so still $R^{'} = 0$ . A nowhere-converging-outside- $c$ series stays that way after differentiating or integrating. (Integrating divides by $n + 1$ , but $1 \cdot \infty = \infty$ too — a melting factor can't cancel an infinite ceiling.)

Extreme $R = \infty$ survives. If the original series is entire ( $lim sup ∣ a_{n} ∣^{1/ n} = 0$ , e.g. $e^{x} = \sum x^{n} / n!$ ), then the differentiated ceiling is $lim sup (n + 1)^{1/ n} ∣ a_{n + 1} ∣^{1/ n} = 1 \cdot 0 = 0$ , so $\frac{1}{R ^{'}} = 0$ and $R^{'} = \infty$ : the differentiated series still converges everywhere. The integration case is identical — the factor $(n + 1)^{- 1/ n} \to 1$ times a ceiling of $0$ is again $0$ , so the antiderivative is entire too. Entire series stay entire under both operations — exactly consistent with $\frac{d}{d x} e^{x} = e^{x}$ and $\int e^{x} d x = e^{x} + C$ , both convergent for every $x$ .

PICTURE. The radius line unchanged for a finite $R$ , with the two red endpoint dots flipping status: one goes ✓→✗ (differentiation), the other ✗→✓ (integration).

The one-picture summary

Here is the entire derivation compressed: the coefficient roots set the ceiling $1/ R$ ; differentiation and integration slap on a factor that the $n$ -th root drives to $1$ ; the ceiling — and the radius — hold firm; only the two endpoints wobble.

Recall Feynman: tell the whole story to a 12-year-old

Picture an endless polynomial living inside a "safe fence" — a distance $R$ left and right of its home base. Inside the fence it adds up to a real function; outside, it explodes.

Now do slope-ing (differentiate): every brick gets multiplied by its step-number $n$ . That sounds scary — bigger bricks! But to test the fence we take the $n$ -th root of each brick, and the $n$ -th root of $n$ crawls all the way down to just $1$ . Multiplying by $1$ changes nothing, so the fence doesn't move. If the fence was zero-width (only the home base worked) it stays zero; if it was infinitely wide (works everywhere) it stays infinite.

Area-ing (integrate) does the opposite — it divides bricks by $n + 1$ — and its $n$ -th root also goes to $1$ . The lone "+C" you add is just a height, not a brick, so it never touches the fence either. Fence unmoved again.

But why is slope-ing-each-piece even allowed? Because inside a slightly smaller fence, all the pieces settle down together in lock-step — one single yardstick controls the whole tail (that's uniform convergence / the $M$ -test, which we secured by comparing to a shrinking geometric series). Only on the fence itself, the two endpoints, do things wobble. Slope-ing can knock an endpoint out; area-ing can pull one back in. So: radius always the same, endpoints always re-checked. That's DIRT.

Recall Why exactly

$n^{1/ n} \to 1$ ? Take logs: $ln (n^{1/ n}) = \frac{l n n}{n}$ . The straight-line $n$ beats the crawling $ln n$ , so the ratio $\to 0$ , so $n^{1/ n} = e^{0} = 1$ . ::: This single limit is the engine that keeps the radius fixed.

Connections

Power series and radius of convergence
Cauchy–Hadamard theorem
Geometric series
Uniform convergence and the Weierstrass M-test
Abel's theorem (endpoint behaviour)
Taylor and Maclaurin series
Solving ODEs with power series

Visual walkthrough — Term-by-term differentiation and integration of power series