Term-by-term differentiation and integration of power series: Question bank

The single rule everything tests: inside the open interval $(c - R, c + R)$ a power series $\sum a_{n} (x - c)^{n}$ acts like an infinite polynomial — differentiate/integrate term by term, and the ==radius $R$ never changes, but the endpoints may== change behaviour. Keep Cauchy–Hadamard theorem and Uniform convergence and the Weierstrass M-test in mind — they are the machinery.

True or false — justify

TF1. Differentiating a power series term by term keeps the radius of convergence the same.

True — the factor

n

contributes

n^{1/ n} \to 1

in the Cauchy–Hadamard

lim sup ∣ a_{n} ∣^{1/ n}

, a subgeometric nudge that vanishes in the

n

-th root, so

R

is untouched.

TF2. Differentiating term by term keeps the interval of convergence the same.

False — the radius is preserved but an endpoint can be lost; e.g. a series convergent at

x = c + R

may diverge there after differentiating.

TF3. Integrating a power series can only shrink or preserve the set of

x

where it converges.

False — integration divides coefficients by

n + 1

, which can gain convergence at an endpoint (this is exactly how

arctan

picks up

x = 1

TF4. Because a power series equals an infinite polynomial, you may always swap

\frac{d}{d x}

and

\sum

with no conditions.

False — the swap is licensed only because the derivative series converges uniformly on compact subintervals of

(c - R, c + R)

; without uniform convergence the swap can fail for general function series.

TF5. If

\sum a_{n} x^{n}

converges only at

x = 0

(so

R = 0

), the term-by-term theorem still tells you how to differentiate it.

False — the theorem requires

R > 0

; with

R = 0

there is no open interval to differentiate on, so nothing is claimed.

TF6. The differentiated series

\sum_{n \geq 1} n a_{n} (x - c)^{n - 1}

and the original series always have the same value at a given interior point.

False — they represent

f^{'}

and

f

, which are generally different functions; only the radius of convergence matches, not the sums.

TF7. A function that is

C^{\infty}

(infinitely differentiable) near

c

must equal its Taylor series there.

False — smoothness gives you all the coefficients, but the series may converge to a different function (e.g.

e^{- 1/ x^{2}}

has all-zero Taylor coefficients yet is nonzero).

TF8. Term-by-term integration of

\sum a_{n} (x - c)^{n}

produces a genuine power series with the same centre

c

True —

C + \sum \frac{a _{n}}{n + 1} (x - c)^{n + 1}

is a power series about

c

, and its radius equals the original

R

TF9. You can differentiate a convergent power series term by term as many times as you like inside

(c - R, c + R)

True — each differentiation returns a power series of the same radius

R > 0

, so the process repeats forever; this is why power-series sums are

C^{\infty}

on the open interval.

TF10. Since the geometric series

\sum x^{n}

diverges at

x = 1

, its term-by-term derivative

\sum n x^{n - 1}

also diverges at

x = 1

True — differentiating never creates endpoint convergence; multiplying terms by

n

makes

x = 1

worse, and indeed

\sum n

diverges.

Spot the error

SE1. "

ln (1 + x) = \sum_{n \geq 0} \frac{( - 1 ) ^{n}}{n + 1} x^{n + 1}

holds for all

x

, since we integrated a series."

Wrong — integration preserves the radius

R = 1

, so the identity holds on

(- 1, 1]

only; at

x = - 1

the series is

- \sum \frac{1}{n + 1}

(harmonic, divergent) and beyond

∣ x ∣ > 1

it fails.

SE2. "The Cauchy–Hadamard limit gives

\frac{1}{R ^{'}} = lim sup ∣ n a_{n} ∣^{1/ n} = lim sup (n^{1/ n} ∣ a_{n} ∣^{1/ n})

, and since

n^{1/ n} \to \infty

, the radius must shrink."

The error is

n^{1/ n} \to \infty

; in fact

n^{1/ n} \to 1

(because

\frac{l n n}{n} \to 0

), so the factor drops out and

R^{'} = R

SE3. "To integrate

\sum a_{n} x^{n}

I write

\sum \frac{a _{n}}{n} x^{n + 1}

The divisor is

n + 1

, not

n

— you raise the power to

n + 1

and divide by that new exponent; dividing by

n

would blow up the

n = 0

term.

SE4. "Differentiating

\sum_{n \geq 0} a_{n} x^{n}

gives

\sum_{n \geq 0} n a_{n} x^{n - 1}

Index should start at

n = 1

— the

n = 0

constant term differentiates to

0

; keeping

n = 0

writes a term

0 \cdot a_{0} x^{- 1}

which is meaningless (

x^{- 1}

is not a power-series term).

SE5. "Because term-by-term differentiation is 'just like polynomials', it works for any infinite sum of functions

\sum f_{n} (x)

Only special families (power series, and more generally uniformly convergent derivative series) qualify; a general

\sum f_{n}

can converge while

\sum f_{n}^{'}

diverges or converges to the wrong thing.

SE6. "The Abel endpoint value tells us the derivative series also converges at that endpoint."

No — Abel's theorem addresses continuity of the sum up to an endpoint where the series converges; it says nothing forcing the differentiated series to converge there.

SE7. "Since

\frac{1}{1 + x ^{2}} = \sum (- 1)^{n} x^{2 n}

for

∣ x ∣ < 1

, integrating gives

arctan x

valid for all real

x

because

arctan

is defined everywhere."

The function

arctan

is defined for all

x

, but its power-series representation only converges for

∣ x ∣ \leq 1

(

R = 1

); the two must not be confused.

Why questions

WHY1. Why does multiplying each coefficient by

n

leave the radius unchanged even though every term gets bigger?

Convergence is geometric, controlled by

∣ a_{n} ∣^{1/ n} ∣ x - c ∣

; the extra factor contributes

n^{1/ n} \to 1

, which is invisible in the

n

-th root, so it can only affect a single endpoint, never the geometric decay rate.

WHY2. Why do we take logarithms to prove

n^{1/ n} \to 1

An expression of the indeterminate form

\infty^{0}

is tamed by logs:

ln (n^{1/ n}) = \frac{l n n}{n} \to 0

because logarithm loses to any positive power of

n

, and then

n^{1/ n} = e^{l n n / n} \to e^{0} = 1

WHY3. Why is uniform convergence — not just pointwise — the property that licenses swapping

\frac{d}{d x}

and

\sum

Uniform convergence of the derivative series lets the limit and the derivative operation exchange without the error accumulating unevenly across

x

; pointwise convergence alone permits pathological failures of that exchange.

WHY4. Why must we always re-test the endpoints after differentiating or integrating?

The theorem guarantees only the open-interval behaviour and the radius; the boundary points

x = c \pm R

are a separate convergence question the theorem deliberately leaves open (integration may gain them, differentiation may lose them).

WHY5. Why does integrating the geometric series

\sum (- 1)^{n} x^{2 n}

gain convergence at

x = 1

Integration divides coefficients by

2 n + 1

, turning

(- 1)^{n}

into

\frac{( - 1 ) ^{n}}{2 n + 1}

— an alternating series with terms decreasing to

0

, so Leibniz's test now gives convergence where the original diverged.

WHY6. Why can a power series be differentiated infinitely many times inside its radius, making its sum

C^{\infty}

Each differentiation yields another power series of the same radius

R > 0

; since this can repeat indefinitely on the same open interval, all derivatives exist there.

WHY7. Why is the constant of integration pinned down by evaluating at the centre (e.g.

x = 0

The antiderivative series has value

C

at the centre where all higher terms vanish, so matching the known value of the target function there (like

arctan 0 = 0

ln 1 = 0

) fixes

C

uniquely.

Edge cases

EC1. What does the theorem say when

R = \infty

(e.g. the series for

e^{x}

Term-by-term operations are valid for all real

x

and stay valid for all

x

— an infinite radius stays infinite, with no endpoints to worry about.

EC2. What happens at the centre

x = c

itself, the degenerate "interval" case?

Every power series converges trivially at

x = c

(all terms after

a_{0}

vanish), and this point is always interior to

(c - R, c + R)

whenever

R > 0

, so nothing special is needed.

EC3. If the original series converges at both endpoints, is the integrated series guaranteed to converge at both?

Not guaranteed as a rule — integration tends to help endpoints (dividing by

n + 1

), so it typically preserves or gains convergence there, but each endpoint must still be checked individually.

EC4. What if

a_{0} \neq = 0

but you only want the derivative — does the constant term matter?

No —

a_{0}

differentiates to

0

and simply drops out; that is why the derivative series starts its index at

n = 1

EC5. Consider a series whose radius is

R = 0

(converges only at the centre). Can you meaningfully integrate it term by term?

No useful function results — with no open interval of convergence there is no interval on which the antiderivative series represents anything, so the theorem does not apply.

EC6. At an endpoint where the differentiated series diverges, is the original function still differentiable there?

The theorem is silent — it only asserts differentiability on the open interval; behaviour of

f

at the boundary requires separate one-sided analysis and is not delivered by the term-by-term rule.

Recall One-line summary you should be able to recite

Inside the open interval you may differentiate and integrate term by term keeping the radius; only the endpoints are up for grabs — differentiation can lose them, integration can gain them. DIRT: Differentiate/Integrate keep the Radius, Test endpoints again.

Connections

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