Power series — centre, radius of convergence, interval of convergence: Worked examples

This page is the exhaustive worked-example companion to the parent topic. The parent gave you the tools (the Ratio Test, the Root Test, the master limsup formula). Here we make sure no situation can surprise you — we list every kind of case the topic can throw, then hit each one with a fully worked example.

The scenario matrix

Each row below is one case class. The rightmost column names the example that lands in that cell.

#	Case class	What is special	Covered by
C1	$R = 0$ (degenerate)	converges only at centre	Ex 1
C2	$R = \infty$	converges everywhere, no endpoints	Ex 2
C3	Finite $R$ , both endpoints diverge	open interval $(a - R, a + R)$	Ex 3
C4	Finite $R$ , one endpoint holds, one breaks	half-open interval	Ex 4
C5	Finite $R$ , both endpoints hold	closed interval $[a - R, a + R]$	Ex 5
C6	Centre $a \neq = 0$ , negative sign in coefficients	shifted, sign bookkeeping	Ex 4, Ex 6
C7	Ratio Test silent, limsup wins	oscillating coefficients	Ex 7
C8	Gap series ( $x^{2 n}$ , only even powers)	substitution trick	Ex 8
C9	Word problem (real-world quantity)	interpret $R$ physically	Ex 9
C10	Exam twist — nested/combined	derivative of a series	Ex 10

The figure below draws exactly those four shapes on one number line. Every row is a finite interval centred at $a$ with the same radius $R$ ; the only difference between rows is the endpoints. A filled dot means "this endpoint is included" (the series converges there); an open (hollow) dot means "excluded" (it diverges there). Read the four coloured bars top to bottom and you have seen every possible outcome a finite radius can produce — the examples on this page each land in one of these rows.

Figure — Power series — centre, radius of convergence, interval of convergence

Case C1 — the radius is zero

Worked example Ex 1 · cell C1 —

$n = 0 \sum \infty n^{n} x^{n}$ Forecast: the coefficients $n^{n}$ grow savagely fast. Do you think this converges for a big set of $x$ , a small set, or almost nothing?

Step 1. Identify centre and coefficients: $a = 0$ , $c_{n} = n^{n}$ . Why this step? Everything downstream measures distance from the centre $a$ ; here the series is written in plain $x^{n}$ , so $a = 0$ .

Step 2. Use the Root Test because $n n^{n}$ simplifies exactly (a coefficient that is itself an $n$ -th power is tailor-made for the $n$ -th root, which cancels the outer power — the Ratio Test would leave a messy $(1 + \frac{1}{n})^{n}$ factor instead). $n ∣ c_{n} ∣ ∣ x ∣^{n} = n n^{n} \cdot ∣ x ∣ = n ∣ x ∣.$ Why this step? $n n^{n} = n$ exactly — the root undoes the $n$ -th power.

Step 3. Take the limit: for any fixed $x \neq = 0$ , $n ∣ x ∣ \to \infty > 1$ , so the series diverges for every $x \neq = 0$ . Why this step? The Root Test says diverge when the limit exceeds $1$ ; here it exceeds $1$ for all $n$ large.

Step 4. At $x = 0$ every term after the first is $0$ , so it converges (to $0^{0} = 1$ by convention). Hence $R = 0$ , interval $= {0}$ .

Verify: $1/ R = lim sup n n^{n} = lim sup n = \infty$ , and $1/\infty$ convention gives $R = 0$ . ✓ Matches.

Case C2 — the radius is infinite

Worked example Ex 2 · cell C2 —

$n = 0 \sum \infty \frac{x ^{n}}{( 2 n )!}$ Forecast: factorials in the denominator crush terms. Guess: finite $R$ or infinite $R$ ?

Step 1. $a = 0$ , $c_{n} = 1/ (2 n)!$ . Use the Ratio Test — factorials cancel beautifully as ratios. Why this step? $\frac{( 2 n )!}{( 2 n + 2 )!}$ is a small finite product, ideal for the ratio.

Step 2. $\frac{a _{n + 1}}{a _{n}} = \frac{∣ x ∣ ^{n + 1} / ( 2 n + 2 )!}{∣ x ∣ ^{n} / ( 2 n )!} = \frac{∣ x ∣}{( 2 n + 1 ) ( 2 n + 2 )} .$ Why this step? $(2 n + 2)! = (2 n + 2) (2 n + 1) (2 n)!$ , so the ratio of factorials is $\frac{1}{( 2 n + 1 ) ( 2 n + 2 )}$ .

Step 3. As $n \to \infty$ this $\to 0 < 1$ for every $x$ . So it converges everywhere: $R = \infty$ . Why this step? Limit is $0$ , always below $1$ , regardless of $x$ — no endpoint check exists (there are no finite endpoints).

Verify: at $x = 1$ the number series $\sum 1/ (2 n)! = cosh (1) \approx 1.5431$ , a finite value — consistent with convergence. ✓ (Writing $x = t^{2}$ shows this is the Maclaurin series of $cosh (t) = cosh (x)$ ; note the argument is $x$ , not $x$ .)

Case C3 — finite $R$ , both endpoints diverge (open interval)

Worked example Ex 3 · cell C3 —

$n = 0 \sum \infty \frac{( x - 5 ) ^{n}}{3 ^{n}}$ Forecast: this is a Geometric Series in disguise, ratio $\frac{x - 5}{3}$ . What is $R$ , and what happens at the two ends?

Step 1. $a = 5$ , $c_{n} = 1/ 3^{n}$ . Rewrite: $\sum (\frac{x - 5}{3})^{n}$ — a geometric series with ratio $r = \frac{x - 5}{3}$ . Why this step? Recognising the Geometric Series form gives the answer instantly: it converges iff $∣ r ∣ < 1$ .

Step 2. $∣ r ∣ < 1 ⟺ ∣ x - 5∣ < 3 \Rightarrow R = 3$ , open interval $(2, 8)$ . Why this step? Geometric convergence is exactly $∣ r ∣ < 1$ ; solving the distance inequality gives the radius.

Step 3. Endpoint $x = 8$ : $r = 1$ , series is $\sum 1^{n} = 1 + 1 + 1 + \dots$ — diverges (terms don't $\to 0$ ). Endpoint $x = 2$ : $r = - 1$ , series is $\sum (- 1)^{n}$ — diverges (terms don't $\to 0$ ). Why this step? At $∣ x - a ∣ = R$ the Ratio Test is silent, so we substitute; both give terms that fail the divergence test.

Step 4. Both ends break $\Rightarrow$ open interval $(2, 8)$ .

Verify: at the midpoint $x = 5$ (centre) sum $= \sum 0^{n} = 1$ , finite ✓. At $x = 6$ , $r = 1/3$ , sum $= \frac{1}{1 - 1/3} = \frac{3}{2}$ , finite ✓.

Case C4 & C6 — one endpoint holds, one breaks; shifted centre

Worked example Ex 4 · cells C4, C6 —

$n = 1 \sum \infty \frac{( - 1 ) ^{n} ( x + 4 ) ^{n}}{n 5 ^{n}}$ Forecast: centre is negative, and there's an alternating sign. Predict the shape: $[)$ , $(]$ , $[]$ or $()$ ?

Step 1. $a = - 4$ (because $x + 4 = x - (- 4)$ ), $c_{n} = \frac{( - 1 ) ^{n}}{n 5 ^{n}}$ . Why this step? Always read the centre as the value of $x$ that makes the base zero: $x + 4 = 0 \Rightarrow x = - 4$ .

Step 2. Ratio Test on $∣ a_{n} ∣ = \frac{∣ x + 4 ∣ ^{n}}{n 5 ^{n}}$ : $\frac{a _{n + 1}}{a _{n}} = \frac{∣ x + 4∣}{5} \cdot \frac{n}{n + 1} \to \frac{∣ x + 4∣}{5} .$ Why this step? The $∣ x + 4 ∣^{n}$ collapses to one factor, $5^{n}$ to a $\frac{1}{5}$ , and $\frac{n}{n + 1} \to 1$ .

Step 3. Converge when $\frac{∣ x + 4∣}{5} < 1 \Rightarrow ∣ x + 4∣ < 5 \Rightarrow R = 5$ , interval $(- 9, 1)$ . Why this step? The Ratio Test guarantees (absolute) convergence exactly when the ratio limit is below $1$ ; setting $\frac{∣ x + 4∣}{5} < 1$ and solving isolates the distance $∣ x + 4∣ < 5$ , whose bound is the radius $R = 5$ and whose solution set is the interval $(a - R, a + R) = (- 9, 1)$ .

Step 4. Endpoint $x = 1$ : $∣ x + 4∣ = 5$ , term $= \frac{( - 1 ) ^{n} 5 ^{n}}{n 5 ^{n}} = \frac{( - 1 ) ^{n}}{n}$ — alternating harmonic, converges by the Alternating Series Test (terms $↓ 0$ ). This is conditional convergence. Endpoint $x = - 9$ : $x + 4 = - 5$ , term $= \frac{( - 1 ) ^{n} ( - 5 ) ^{n}}{n 5 ^{n}} = \frac{( - 1 ) ^{n} ( - 1 ) ^{n}}{n} = \frac{1}{n}$ — the harmonic p-series with $p = 1$ , diverges. Why this step? The two endpoints are genuinely different series; you must substitute each and pick the right test.

Step 5. Left breaks, right holds $\Rightarrow (- 9, 1]$ .

Verify: $x = - 9$ collapses the two signs $(- 1)^{n} (- 1)^{n} = 1$ , giving the divergent harmonic series ✓. $x = 1$ gives $\sum (- 1)^{n} / n = - ln 2 \approx - 0.693$ , finite ✓.

Case C5 — both endpoints hold (closed interval)

Worked example Ex 5 · cell C5 —

$n = 1 \sum \infty \frac{( x - 1 ) ^{n}}{n ^{2}}$ Forecast: the $n^{2}$ in the denominator is strong. Guess whether both endpoints survive.

Step 1. $a = 1$ , $c_{n} = 1/ n^{2}$ . Use the Ratio Test here: the coefficient is a simple power $n^{- 2}$ , so the ratio $\frac{c _{n + 1}}{c _{n}} = \frac{n ^{2}}{( n + 1 ) ^{2}}$ is a clean rational function with an obvious limit — far easier than taking $n 1/ n^{2}$ , which needs the $n^{1/ n} \to 1$ fact. $\frac{a _{n + 1}}{a _{n}} = ∣ x - 1∣ \cdot \frac{n ^{2}}{( n + 1 ) ^{2}} \to ∣ x - 1∣.$ Why this step? $\frac{n ^{2}}{( n + 1 ) ^{2}} \to 1$ , so the whole limit is just $∣ x - 1∣$ .

Step 2. Converge when $∣ x - 1∣ < 1 \Rightarrow R = 1$ , interval $(0, 2)$ .

Step 3. Endpoint $x = 2$ : $\sum 1/ n^{2}$ — p-series with $p = 2 > 1$ , converges. Endpoint $x = 0$ : $\sum (- 1)^{n} / n^{2}$ — converges absolutely (its absolute version is the same $p = 2$ series) by absolute convergence. Why this step? When $p > 1$ the p-series holds; the sign in front cannot break an already-absolutely-convergent series.

Step 4. Both hold $\Rightarrow [0, 2]$ .

Verify: $x = 2$ gives $\sum 1/ n^{2} = π^{2} /6 \approx 1.6449$ , finite ✓. $x = 0$ gives $\sum (- 1)^{n} / n^{2} = - π^{2} /12 \approx - 0.8225$ , finite ✓.

Case C6 (revisited) — sign inside the base, careful bookkeeping

Worked example Ex 6 · cell C6 —

$n = 0 \sum \infty \frac{( 3 - x ) ^{n}}{2 ^{n} ( n + 1 )}$ Forecast: the base is $3 - x$ , not $x - 3$ . Does that flip anything? Where is the centre?

Step 1. Note $3 - x = - (x - 3)$ , so $(3 - x)^{n} = (- 1)^{n} (x - 3)^{n}$ . Centre $a = 3$ . Why this step? Factoring out $(- 1)^{n}$ shows the true centre and reveals a hidden alternating sign at one endpoint — a classic trap.

Step 2. Ratio Test on $∣ a_{n} ∣ = \frac{∣ x - 3 ∣ ^{n}}{2 ^{n} ( n + 1 )}$ : $\frac{a _{n + 1}}{a _{n}} = \frac{∣ x - 3∣}{2} \cdot \frac{n + 1}{n + 2} \to \frac{∣ x - 3∣}{2} .$ Step 3. $\frac{∣ x - 3∣}{2} < 1 \Rightarrow R = 2$ , interval $(1, 5)$ . Why this step? The Ratio Test delivers convergence precisely when its limit is less than $1$ ; the inequality $\frac{∣ x - 3∣}{2} < 1$ rearranges to the distance condition $∣ x - 3∣ < 2$ , whose bound is the radius $R = 2$ and whose solution is the interval $(a - R, a + R) = (1, 5)$ .

Step 4. Endpoint $x = 5$ : $3 - x = - 2$ , term $= \frac{( - 2 ) ^{n}}{2 ^{n} ( n + 1 )} = \frac{( - 1 ) ^{n}}{n + 1}$ — alternating, converges (Alternating Series Test). Endpoint $x = 1$ : $3 - x = 2$ , term $= \frac{2 ^{n}}{2 ^{n} ( n + 1 )} = \frac{1}{n + 1}$ — harmonic (shifted), diverges.

Step 5. $(1, 5]$ .

Verify: $x = 1$ gives $\sum \frac{1}{n + 1}$ = harmonic tail, diverges ✓. $x = 5$ gives $\sum \frac{( - 1 ) ^{n}}{n + 1} = ln 2 \approx 0.693$ , finite ✓.

Case C7 — the Ratio Test is silent, limsup wins

Worked example Ex 7 · cell C7 — coefficients

$c_{n} = {3^{n} 1 n even n odd$ , series $\sum c_{n} x^{n}$ Forecast: the ratio $∣ c_{n + 1} / c_{n} ∣$ will swing wildly. Can the Ratio Test give $R$ here at all?

Step 1. Compute the ratio: from even→odd it's $\frac{1}{3 ^{n}}$ (tiny), from odd→even it's $3^{n + 1}$ (huge). No limit exists. Why this step? This demonstrates the Ratio formula's failure mode directly.

Step 2. Switch to the master root/limsup tool (the Cauchy–Hadamard formula above), which always exists: $n ∣ c_{n} ∣ = {n 3^{n} = 3 n 1 = 1 n even n odd$ Why this step? The $n$ -th root converts the $3^{n}$ into a constant $3$ , taming the oscillation.

Step 3. The values cluster at ${1, 3}$ ; the largest cluster point is $lim sup n ∣ c_{n} ∣ = 3$ . Why this step? The limsup is the largest limit of any subsequence — here $3$ .

Step 4. Cauchy–Hadamard: $R = \frac{1}{lim sup n ∣ c _{n} ∣} = \frac{1}{3}$ , open-so-far interval $(- \frac{1}{3}, \frac{1}{3})$ .

Step 5. Endpoints $x = \pm \frac{1}{3}$ : look at the even terms alone, $c_{2 k} x^{2 k} = 3^{2 k} x^{2 k} = (3 x)^{2 k}$ . At $x = \pm \frac{1}{3}$ this is $(\pm 1)^{2 k} = 1$ for every $k$ , so infinitely many terms equal $1$ and do not tend to $0$ . The divergence test fails ⇒ the series diverges at both endpoints. Why this step? At $∣ x ∣ = R$ neither test formula applies, so we inspect the terms directly; a subsequence of terms staying at $1$ already kills convergence.

Step 6. Both ends break $\Rightarrow$ interval $(- \frac{1}{3}, \frac{1}{3})$ .

Verify: $1/ R = 3 \Rightarrow R = 1/3$ . Sanity: the even terms $3^{n} x^{n} = (3 x)^{n}$ form a geometric part needing $∣3 x ∣ < 1 ⟺ ∣ x ∣ < 1/3$ — exactly $R = 1/3$ , and at $∣ x ∣ = 1/3$ that geometric part has ratio $∣3 x ∣ = 1$ , diverging. ✓

Case C8 — gap series (only even powers)

Worked example Ex 8 · cell C8 —

$n = 1 \sum \infty \frac{x ^{2 n}}{n 9 ^{n}}$ Forecast: there is no $x^{2 n + 1}$ term. Does $R$ come out as $∣ x ∣ < something$ or $x^{2} < something$ ?

Step 1. Substitute $u = x^{2}$ . Then the series is $\sum \frac{u ^{n}}{n 9 ^{n}}$ , an ordinary power series in $u$ . Why this step? Gap series in $x^{2 n}$ become standard series once you rename $x^{2}$ as a single variable.

Step 2. Ratio Test in $u$ : $\frac{a _{n + 1}}{a _{n}} = \frac{∣ u ∣}{9} \cdot \frac{n}{n + 1} \to \frac{∣ u ∣}{9}$ . Converge when $∣ u ∣ < 9$ . Why this step? The $u^{n}$ cancels to one factor, $9^{n}$ to $\frac{1}{9}$ , and $\frac{n}{n + 1} \to 1$ ; the Ratio Test converges when this limit is below $1$ , which is exactly $∣ u ∣ < 9$ .

Step 3. Translate back: $∣ u ∣ = ∣ x^{2} ∣ = x^{2} < 9 \Rightarrow ∣ x ∣ < 3$ . So $R = 3$ , interval $(- 3, 3)$ in $x$ . Why this step? $x^{2} < 9 ⟺ ∣ x ∣ < 3$ ; the radius in $x$ is the square root of the radius in $u$ .

Step 4. Endpoints $x = \pm 3$ : then $x^{2} = 9$ , term $= \frac{9 ^{n}}{n 9 ^{n}} = \frac{1}{n}$ — harmonic, diverges at both ends. Why this step? Even powers make both endpoints give the same value $x^{2} = 9$ , so they behave identically.

Step 5. $(- 3, 3)$ .

Verify: $x^{2} = 9$ gives $\sum 1/ n$ , diverges ✓. At $x = 1$ : $\sum \frac{1}{n 9 ^{n}} = - ln (1 - 1/9) = ln (9/8) \approx 0.1178$ , finite ✓.

Case C9 — a real-world word problem

Worked example Ex 9 · cell C9 — a settling deposit

A financial model pays, in year $n$ , an amount proportional to $\frac{r ^{n}}{n}$ where $r = \frac{x - 100}{20}$ measures how far the interest offset $x$ (in the same units) sits from the baseline $100$ . The total value is $V (x) = n = 1 \sum \infty \frac{( x - 100 ) ^{n}}{n 2 0 ^{n}}$ . For which $x$ does the total stay finite (the model doesn't blow up)? Forecast: the "safe operating range" is an interval around $x = 100$ . Guess its width.

Step 1. $a = 100$ , $c_{n} = 1/ (n 2 0^{n})$ . Ratio Test: $\frac{∣ x - 100∣}{20} \cdot \frac{n}{n + 1} \to \frac{∣ x - 100∣}{20}$ . Why this step? Same mechanics as Ex 4/6; the offset centre is $100$ .

Step 2. Converge when $∣ x - 100∣ < 20 \Rightarrow R = 20$ , interval $(80, 120)$ . Why this step? $R = 20$ is the maximum offset from the baseline $x = 100$ before the total diverges — the model's tolerance.

Step 3. Endpoint $x = 120$ : $\sum \frac{2 0 ^{n}}{n 2 0 ^{n}} = \sum \frac{1}{n}$ — diverges (model blows up exactly at the top edge). Endpoint $x = 80$ : $\sum \frac{( - 20 ) ^{n}}{n 2 0 ^{n}} = \sum \frac{( - 1 ) ^{n}}{n}$ — alternating, converges (bottom edge is safe).

Step 4. Safe range $[80, 120)$ : the offset may go down by 20 (inclusive) but must stay strictly below +20.

Verify: $x = 80$ gives $\sum (- 1)^{n} / n = - ln 2 \approx - 0.693$ , finite ✓. $x = 120$ gives harmonic, diverges ✓. Radius $= 20$ matches $∣ c_{n} / c_{n + 1} ∣ = \frac{( n + 1 ) 2 0 ^{n + 1}}{n 2 0 ^{n}} \to 20$ ✓.

Case C10 — exam twist: differentiate a series

Worked example Ex 10 · cell C10 —

$f (x) = n = 1 \sum \infty \frac{x ^{n}}{n}$ ; find $R$ of $f^{'} (x)$ Forecast: Term-by-term differentiation changes the coefficients. Does it change the radius?

Step 1. $f$ has $a = 0$ , $c_{n} = 1/ n$ ; its Ratio Test radius is $R = lim ∣ c_{n} / c_{n + 1} ∣ = lim \frac{n + 1}{n} = 1$ . Why this step? Establish the original radius so we can compare.

Step 2. Differentiate term by term: $f^{'} (x) = \sum_{n = 1}^{\infty} \frac{n x ^{n - 1}}{n} = \sum_{n = 1}^{\infty} x^{n - 1} = \sum_{m = 0}^{\infty} x^{m}$ . Why this step? $\frac{d}{d x} \frac{x ^{n}}{n} = x^{n - 1}$ ; reindex $m = n - 1$ to read off the new series — a plain Geometric Series.

Step 3. The derivative is geometric with ratio $x$ , so its radius is also $R = 1$ . Why this step? The theorem guarantees the radius is unchanged by differentiation — endpoints may change but $R$ does not.

Step 4. Endpoint check (they DO differ): $f$ itself has interval $[- 1, 1)$ ( $x = - 1$ converges, $x = 1$ harmonic diverges). But $f^{'} = \sum x^{m}$ diverges at both $x = \pm 1$ , interval $(- 1, 1)$ . Why this step? Same radius, narrower interval — a favourite exam gotcha.

Verify: $f^{'} (x) = \frac{1}{1 - x}$ (geometric sum) for $∣ x ∣ < 1$ , and indeed $\frac{d}{d x} (- ln (1 - x)) = \frac{1}{1 - x}$ — consistent, since $\sum x^{n} / n = - ln (1 - x)$ . ✓ Radius $1$ for both ✓.

Recall

Recall Which shape? Match series to interval

$\sum \frac{( x - 2 ) ^{n}}{n ^{2}}$ → interval shape? ::: Closed $[1, 3]$ — $p = 2$ holds at both ends. $\sum \frac{( x - 2 ) ^{n}}{3 ^{n}}$ → interval shape? ::: Open $(- 1, 5)$ — geometric, both ends give $\sum (\pm 1)^{n}$ , diverge. Ratio $∣ c_{n + 1} / c_{n} ∣$ oscillates — which tool? ::: The root/limsup Cauchy–Hadamard formula $1/ R = lim sup n ∣ c_{n} ∣$ . A gap series $\sum a_{n} x^{2 n}$ with $u$ -radius $9$ — radius in $x$ ? ::: $9 = 3$ . Does differentiating term-by-term change $R$ ? ::: No — $R$ is preserved; only endpoints may change.