Term-by-term differentiation and integration of power series: Worked examples

Before we compute anything, let me name the four tools so no symbol appears unearned.

The scenario matrix

Every example below is tagged with the cell(s) it covers. The goal: no empty cell. Note the two-sided rows: every "endpoint" cell is checked at both $x = c + R$ and $x = c - R$ .

Cell	The scenario it stress-tests	Covered by
A	Differentiate — all coefficients positive	Ex 1
B	Integrate — series gains the right endpoint $x = + 1$	Ex 2
B′	Integrate — the left endpoint $x = - 1$ (opposite sign)	Ex 2, step 4
C	Integrate — series only conditionally convergent at endpoint	Ex 3
C′	Integrate — the left endpoint $x = - 1$ diverges	Ex 3, step 4
D	Differentiate — series loses an endpoint (both sides)	Ex 4
E	Shifted centre $c \neq = 0$ (not centred at zero)	Ex 5
F	Degenerate / limiting: the series that returns itself	Ex 6
G	Real-world word problem (physics)	Ex 7
H	Exam twist: sum an unknown series by recognising a derivative	Ex 8

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

The figure above is the map: the horizontal line is the $x$ -axis, the shaded band is the safe zone $(c - R, c + R)$ , and the two dots are the endpoints — the only places where differentiating or integrating can flip the verdict. Both dots must be tested independently.

Cell A — differentiate a positive series

Worked example Example 1 — from

$\frac{1}{1 - x}$ to $\frac{1}{( 1 - x ) ^{2}}$ and beyond Start from the Geometric series $\frac{1}{1 - x} = n = 0 \sum \infty x^{n}$ for $∣ x ∣ < 1$ . Find a closed form for $n = 1 \sum \infty n^{2} x^{n}$ .

Forecast: guess before reading — differentiating once gave $\frac{1}{( 1 - x ) ^{2}}$ ; a second differentiation (with a multiply-by- $x$ trick) should produce a rational function with a cube in the denominator. Write down your guess.

Differentiate once. $\frac{d}{d x} \frac{1}{1 - x} = \frac{1}{( 1 - x ) ^{2}} = \sum_{n = 1}^{\infty} n x^{n - 1}$ . Why this step? The theorem lets each term $x^{n}$ become $n x^{n - 1}$ ; the $n = 0$ term is a constant so it vanishes, which is why the sum now starts at $n = 1$ .
Multiply by $x$ to line the powers back up: $\frac{x}{( 1 - x ) ^{2}} = \sum_{n = 1}^{\infty} n x^{n}$ . Why this step? We want a $\sum n^{2} x^{n}$ eventually; keeping the exponent equal to $n$ (not $n - 1$ ) means the next differentiation will hand us another factor of $n$ cleanly.
Differentiate again. $\frac{d}{d x} \frac{x}{( 1 - x ) ^{2}} = \frac{1 + x}{( 1 - x ) ^{3}} = \sum_{n = 1}^{\infty} n^{2} x^{n - 1}$ . Why this step? Differentiating $n x^{n}$ gives $n \cdot n x^{n - 1} = n^{2} x^{n - 1}$ — the second factor of $n$ we wanted.
Multiply by $x$ once more. $n = 1 \sum \infty n^{2} x^{n} = \frac{x ( 1 + x )}{( 1 - x ) ^{3}}$ .

Verify: put $x = \frac{1}{2}$ . Closed form: $\frac{\frac{1}{2} \cdot \frac{3}{2}}{( \frac{1}{2} ) ^{3}} = \frac{3/4}{1/8} = 6$ . Partial sum $\sum_{n = 1}^{6} n^{2} (\frac{1}{2})^{n} = 0.5 + 1 + 1.125 + 1 + 0.78125 + 0.5625 = 4.968 \dots$ , marching toward $6$ . ✓ Radius unchanged at $R = 1$ (each differentiation kept it — radius-survival).

Cell B / B′ — integration gains an endpoint (both sides checked)

Worked example Example 2 — the

$arctan$ series and Leibniz's $π$ Derive the series for $arctan x$ and show what happens at both endpoints $x = 1$ and $x = - 1$ .

Forecast: the base series $\frac{1}{1 + x ^{2}}$ diverges at $x = \pm 1$ (terms $\pm 1$ never shrink). After integrating, do you expect $x = \pm 1$ to still fail, or to suddenly work? Guess.

Match the geometric pattern. $\frac{1}{1 + x ^{2}} = \frac{1}{1 - ( - x ^{2} )} = \sum_{n = 0}^{\infty} (- x^{2})^{n} = \sum_{n = 0}^{\infty} (- 1)^{n} x^{2 n}$ , valid when $∣ - x^{2} ∣ < 1$ , i.e. $∣ x ∣ < 1$ . Why this step? Substituting $- x^{2}$ for the " $x$ " of the geometric series is the cheapest way to get a series for $\frac{1}{1 + x ^{2}}$ — no calculus needed yet.
Integrate from $0$ to $x$ , term by term: $arctan x = \int_{0}^{x} \frac{d t}{1 + t ^{2}} = n = 0 \sum \infty \frac{( - 1 ) ^{n}}{2 n + 1} x^{2 n + 1}$ . Why this step? $\int_{0}^{x} t^{2 n} d t = \frac{x ^{2 n + 1}}{2 n + 1}$ , and the lower limit $0$ auto-fixes the constant since $arctan 0 = 0$ .
Right endpoint $x = 1$ — two-step. (1) AST: the terms become $1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots$ , which alternate and shrink to $0$ , so by the alternating series test the series converges. (2) Abel: since it converges, Abel's theorem identifies the sum as $lim_{x \to 1^{-}} arctan x = arctan 1 = \frac{π}{4}$ . Why this step? Keep the two tools apart: AST establishes convergence, Abel pins the value. Dividing by $2 n + 1$ shrank the terms enough — the right endpoint was gained by integrating.
Left endpoint $x = - 1$ — the opposite sign, checked separately. Plugging $x = - 1$ : the odd power $x^{2 n + 1} = - 1$ gives $\sum \frac{( - 1 ) ^{n} ( - 1 )}{2 n + 1} = - (1 - \frac{1}{3} + \frac{1}{5} - \dots)$ . (1) AST: same alternating, shrinking terms → converges. (2) Abel: sum $= arctan (- 1) = - \frac{π}{4}$ . Why this step? The theorem demands each boundary be tested on its own. Here $arctan$ is odd, so the left endpoint mirrors the right — but you only know that by doing the check, not assuming it.

Verify: $4 \sum_{n = 0}^{5000} \frac{( - 1 ) ^{n}}{2 n + 1} \approx 3.1414$ , closing on $π \approx 3.14159$ (slowly, oscillating). And at $x = - 1$ : $\sum_{n = 0}^{5000} \frac{( - 1 ) ^{n} ( - 1 )}{2 n + 1} \approx - 0.7854 \approx - \frac{π}{4}$ . ✓ Radius still $R = 1$ .

The figure shows why the endpoints are gained: the base series' terms (yellow) sit at height $1$ forever, but after integrating, the new terms (blue) decay like $\frac{1}{2 n + 1} \to 0$ — the same picture at $x = 1$ and $x = - 1$ .

Cell C / C′ — integration to a conditionally convergent endpoint (both sides)

Worked example Example 3 —

$ln (1 + x)$ and the alternating harmonic series From $\frac{1}{1 + x} = \sum_{n = 0}^{\infty} (- 1)^{n} x^{n}$ , build $ln (1 + x)$ and evaluate at both endpoints $x = 1$ and $x = - 1$ .

Forecast: will the endpoint sums converge absolutely (rearrange freely) or only conditionally (order matters)? And will both sides behave the same? Decide before steps 3–4.

Integrate the base series $0 \to x$ : $ln (1 + x) = \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n}}{n + 1} x^{n + 1} = x - \frac{x ^{2}}{2} + \frac{x ^{3}}{3} - \dots$ . Why this step? $\int_{0}^{x} t^{n} d t = \frac{x ^{n + 1}}{n + 1}$ ; constant is $0$ because $ln (1 + 0) = ln 1 = 0$ .
The safe zone is $∣ x ∣ < 1$ , so the two endpoints to test are $x = 1$ and $x = - 1$ . Why this step? Base series has $R = 1$ ; integrating keeps $R = 1$ (radius-survival), so the boundaries are exactly $\pm 1$ .
Right endpoint $x = 1$ — two-step. (1) AST: $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots$ alternates and shrinks to $0$ → converges. (2) Abel: sum $= lim_{x \to 1^{-}} ln (1 + x) = ln 2$ . Absolute version $1 + \frac{1}{2} + \frac{1}{3} + \dots$ is the harmonic series, which diverges — so the right endpoint is only conditionally convergent. Why this step? Separating AST (converges) from absolute convergence (fails) is exactly what "conditional" means.
Left endpoint $x = - 1$ — checked separately, and it FAILS. Plugging $x = - 1$ : $x^{n + 1} = (- 1)^{n + 1}$ gives $\sum \frac{( - 1 ) ^{n} ( - 1 ) ^{n + 1}}{n + 1} = \sum \frac{( - 1 ) ^{2 n + 1}}{n + 1} = - \sum_{n = 0}^{\infty} \frac{1}{n + 1}$ — the negative harmonic series, which diverges to $- \infty$ . This matches $ln (1 + (- 1)) = ln 0 = - \infty$ . Why this step? This is the crucial asymmetry: the right endpoint was gained but the left endpoint is lost. You would never see it if you only tested $x = + 1$ . Always do both.

Verify: $\sum_{n = 0}^{9999} \frac{( - 1 ) ^{n}}{n + 1} \approx 0.69310$ , and $ln 2 \approx 0.69315$ . Left side $\sum_{n = 0}^{9999} \frac{1}{n + 1} \approx 9.79$ and still climbing (diverges). ✓ Radius $R = 1$ .

Cell D — differentiation loses an endpoint (both sides)

Worked example Example 4 — where differentiating breaks the edge

Consider $f (x) = n = 1 \sum \infty \frac{x ^{n}}{n ^{2}}$ . Its radius is $R = 1$ , and at both $x = 1$ and $x = - 1$ it converges (at $x = 1$ it is $\sum \frac{1}{n ^{2}} = \frac{π ^{2}}{6}$ ; at $x = - 1$ it is $\sum \frac{( - 1 ) ^{n}}{n ^{2}}$ , absolutely convergent). What happens to the endpoints after differentiating?

Forecast: the coefficients are $\frac{1}{n ^{2}}$ (fast decay, works at both $\pm 1$ ). Differentiating multiplies by $n$ . Do you predict $x = 1$ and $x = - 1$ survive?

Differentiate term by term: $f^{'} (x) = \sum_{n = 1}^{\infty} \frac{n x ^{n - 1}}{n ^{2}} = \sum_{n = 1}^{\infty} \frac{x ^{n - 1}}{n}$ . Why this step? $\frac{d}{d x} \frac{x ^{n}}{n ^{2}} = \frac{n x ^{n - 1}}{n ^{2}} = \frac{x ^{n - 1}}{n}$ : the factor $n$ from differentiating cancelled one power of $n$ in the denominator, leaving decay only like $\frac{1}{n}$ .
Right endpoint $x = 1$ : $\sum_{n = 1}^{\infty} \frac{1}{n}$ — the harmonic series, which diverges. Right endpoint lost. Why this step? Before differentiating $\sum \frac{1}{n ^{2}}$ converged; after, $\sum \frac{1}{n}$ diverges.
Left endpoint $x = - 1$ — checked separately. Here $x^{n - 1} = (- 1)^{n - 1}$ gives $\sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n - 1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \dots$ , which by AST converges (to $ln 2$ ). So the left endpoint survives even though the right one died. Why this step? This is the sharpest possible reminder: the two endpoints can have opposite fates. Differentiation lost $x = 1$ but kept $x = - 1$ . Testing only one side would give a false conclusion.

Verify: the derivative is $- \frac{l n ( 1 - x )}{x}$ in closed form; near $x = 1^{-}$ it blows up like $- ln (1 - x) \to + \infty$ (right endpoint gone), while at $x = - 1$ it equals $- \frac{l n 2}{- 1} = ln 2 \approx 0.693$ (finite, left endpoint kept). Numerically at $x = 0.99$ : $\sum_{n = 1}^{2000} \frac{0.9 9 ^{n - 1}}{n} \approx 4.20$ , and $- ln (0.01) /0.99 \approx 4.65$ — growing. ✓ Radius still $R = 1$ .

Cell E — a shifted centre $c \neq = 0$

Worked example Example 5 — series centred at

$c = 2$ Let $g (x) = n = 0 \sum \infty \frac{( x - 2 ) ^{n}}{3 ^{n}}$ . Find $g^{'} (x)$ as a series and as a closed form, and state its safe zone.

Forecast: the " $x$ " of the geometric pattern is now $\frac{x - 2}{3}$ . The centre is $2$ , and $R$ comes from $\frac{x - 2}{3} < 1$ . Guess $R$ .

Recognise the geometric shape. With $u = \frac{x - 2}{3}$ , $g = \sum u^{n} = \frac{1}{1 - u} = \frac{1}{1 - \frac{x - 2}{3}} = \frac{3}{5 - x}$ , valid for $∣ u ∣ < 1 \Rightarrow ∣ x - 2∣ < 3$ , so $c = 2, R = 3$ . Why this step? Naming $u$ makes the shift invisible — it's still just a geometric series in disguise.
Differentiate the series. $g^{'} (x) = \sum_{n = 1}^{\infty} \frac{n ( x - 2 ) ^{n - 1}}{3 ^{n}}$ . Why this step? The theorem is centre-agnostic: $\frac{d}{d x} (x - 2)^{n} = n (x - 2)^{n - 1}$ exactly as if $c$ were $0$ .
Differentiate the closed form to cross-check. $g^{'} (x) = \frac{d}{d x} \frac{3}{5 - x} = \frac{3}{( 5 - x ) ^{2}}$ . Why this step? Two independent routes must agree — that's how you trust term-by-term work.

Verify: at $x = 3$ (inside, since $∣3 - 2∣ = 1 < 3$ ): series $\sum_{n = 1}^{20} \frac{n \cdot 1 ^{n - 1}}{3 ^{n}} \approx 0.75$ ; closed form $\frac{3}{( 5 - 3 ) ^{2}} = \frac{3}{4} = 0.75$ . ✓ Centre $2$ , radius $3$ both unchanged by differentiation. The endpoints here are $x = - 1$ and $x = 5$ ; both diverge (geometric terms $\pm 1$ don't shrink), on both sides.

Cell F — the degenerate / self-returning case

Worked example Example 6 — the series equal to its own derivative

Show $E (x) = n = 0 \sum \infty \frac{x ^{n}}{n !}$ satisfies $E^{'} = E$ , and note why this is the limiting/degenerate case where nothing at all changes.

Forecast: the exponential Taylor and Maclaurin series has $R = \infty$ (no endpoints exist). Predict what $E^{'}$ looks like.

Differentiate. $E^{'} (x) = \sum_{n = 1}^{\infty} \frac{n x ^{n - 1}}{n !} = \sum_{n = 1}^{\infty} \frac{x ^{n - 1}}{( n - 1 )!}$ . Why this step? $\frac{n}{n !} = \frac{1}{( n - 1 )!}$ — the factor $n$ from differentiating cancels the top of the factorial.
Re-index $m = n - 1$ : $E^{'} (x) = \sum_{m = 0}^{\infty} \frac{x ^{m}}{m !} = E (x)$ . Why this step? Shifting the index shows the differentiated series is identical to the original.
Note the degeneracy. Here $R = \infty$ , so the safe zone is the whole line and there are no endpoints to gain or lose — neither $+ \infty$ nor $- \infty$ is a real boundary point. This is the one cell where the "endpoint caveat" is vacuously satisfied.

Verify: at $x = 1$ , $\sum_{n = 0}^{10} \frac{1}{n !} \approx 2.71828$ , matching $e$ . Derivative sum equals it by construction. ✓

Cell G — a real-world word problem

Worked example Example 7 — charging a capacitor, term by term (with units)

In an RC circuit a capacitor charges so its voltage is $V (t) = V_{0} (1 - e^{- t / τ})$ , where $V_{0}$ is measured in volts (V), $t$ in seconds (s), and $τ$ (the time constant) also in seconds. The current is $I (t) = C V^{'} (t)$ where $C$ is capacitance in farads (F). Find $V (t)$ as a power series in $t$ and differentiate it to get $I (t)$ , checking units.

Forecast: $e^{- t / τ}$ near $t = 0$ is roughly $1 - \frac{t}{τ}$ , so $V \approx V_{0} \frac{t}{τ}$ (linear ramp at first). Predict the leading term of $I$ .

Expand the exponential. $e^{- t / τ} = \sum_{n = 0}^{\infty} \frac{( - t / τ ) ^{n}}{n !}$ , so $V (t) = V_{0} (1 - \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n} t ^{n}}{n ! τ ^{n}}) = V_{0} \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n - 1} t ^{n}}{n ! τ ^{n}}$ . Why this step? The $n = 0$ term of $e^{- t / τ}$ is $1$ , and $1 - 1 = 0$ , so the series starts at $n = 1$ : a clean ramp with no constant offset — matching $V (0) = 0$ .
Differentiate term by term. $V^{'} (t) = V_{0} \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n - 1} n t ^{n - 1}}{n ! τ ^{n}} = V_{0} \sum_{n = 1}^{\infty} \frac{( - 1 ) ^{n - 1} t ^{n - 1}}{( n - 1 )! τ ^{n}} = \frac{V _{0}}{τ} e^{- t / τ}$ . Why this step? Same $\frac{n}{n !}$ cancellation as Example 6; re-summing gives the compact closed form.
Multiply by $C$ . $I (t) = C V^{'} (t) = \frac{C V _{0}}{τ} e^{- t / τ}$ , and at $t = 0$ : $I (0) = \frac{C V _{0}}{τ}$ . Why this step? This is the peak initial current; term-by-term differentiation handed us the whole current curve.

Verify (units). $\frac{C V _{0}}{τ}$ has units $\frac{F \cdot V}{s}$ . Since $1 F = 1 C / V$ (coulomb per volt), $\frac{F \cdot V}{s} = \frac{C}{s} = A$ (amperes). ✓ Current in amps — dimensionally correct. Numerically with $V_{0} = 5$ V, $τ = 2$ s, $C = 1$ F, $t = 1$ s: $I = \frac{5}{2} e^{- 0.5} \approx 1.516$ A.

Cell H — the exam twist

Worked example Example 8 — sum a mystery series by spotting a derivative

Evaluate $S = n = 1 \sum \infty \frac{n}{2 ^{n}}$ in closed form.

Forecast: there's an $n$ upstairs — a tell-tale sign a differentiation created it. Guess whether $S$ is a small integer.

Find the parent series. Recall $\sum_{n = 1}^{\infty} n x^{n - 1} = \frac{1}{( 1 - x ) ^{2}}$ (Example 1, step 1). Why this step? The factor $n$ in $\frac{n}{2 ^{n}}$ screams "I came from differentiating $\sum x^{n}$ ."
Multiply by $x$ to match the exponent: $\sum_{n = 1}^{\infty} n x^{n} = \frac{x}{( 1 - x ) ^{2}}$ . Why this step? Our target has $x^{n}$ (namely $x = \frac{1}{2}$ so $x^{n} = 1/ 2^{n}$ ), not $x^{n - 1}$ .
Plug $x = \frac{1}{2}$ (safely inside $R = 1$ ): $S = \sum_{n = 1}^{\infty} n (\frac{1}{2})^{n} = \frac{\frac{1}{2}}{( 1 - \frac{1}{2} ) ^{2}} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2$ . Why this step? Since $∣ x ∣ = \frac{1}{2} < 1$ sits strictly inside the safe zone, no endpoint subtlety arises — the term-by-term-derived closed form is valid, and it gives the final answer $S = 2$ .

Verify: partial sum $\sum_{n = 1}^{20} \frac{n}{2 ^{n}} \approx 1.99998$ , converging to $2$ . ✓ The final closed form is $S = 2$ .

Recall Which cell was hardest for you?

Cell D (differentiation losing an endpoint) vs Cell B (integration gaining one) — can you state the one-sentence reason each happens, and why you must check both sides? Losing: differentiating multiplies coefficients by $n$ , weakening decay, so a borderline endpoint can break. Gaining: integrating divides by $n + 1$ , strengthening decay, so a borderline endpoint can start converging. ::: And Ex 3/Ex 4 show the two ends can even disagree — so test $x = c + R$ and $x = c - R$ separately, every time.

Recap

Recall The scenario matrix, from memory

Name all cells A–H and the endpoint rule they enforce. A differentiate-positive; B/B′ integrate-gains-endpoint (both sides); C/C′ integrate-conditional (right gained, left lost); D differentiate-loses-endpoint (right lost, left kept); E shifted-centre; F self-returning/degenerate; G word-problem; H exam-twist. ::: In every one, the radius $R$ was unchanged — only endpoints ever flipped, and each of the two endpoints must be tested on its own.

Connections

Parent topic
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

Worked examples — Term-by-term differentiation and integration of power series