Power series — centre, radius of convergence, interval of convergence: Exercises

Intuition How to use this page

Each problem below asks you to find the radius of convergence $R$ and the full interval of convergence for a power series. Work each one yourself on paper FIRST, then open the collapsible solution. The problems climb five levels: L1 Recognition (read off the answer), L2 Application (run the standard recipe), L3 Analysis (endpoints get subtle), L4 Synthesis (combine several ideas), L5 Mastery (the $lim sup$ master tool where the ratio fails).

The recipe we lean on throughout comes from the parent topic:

Run the Ratio Test (or Root Test) on $∣ c_{n} (x - a)^{n} ∣$ to get a condition of the form $∣ x - a ∣ \cdot k < 1$ . Here == $k$ is the coefficient limit left over after the $∣ x - a ∣$ factor pulls out== — precisely $k = lim_{n \to \infty} ∣ c_{n + 1} / c_{n} ∣$ when you use the Ratio Test, or $k = lim sup_{n \to \infty} n ∣ c_{n} ∣$ when you use the Root Test. Either way $R = 1/ k$ .
Read off $R$ and the open interval $(a - R, a + R)$ .
Substitute each endpoint and test the resulting number series separately.
Write the interval, including or excluding each endpoint.

Before we start, one picture to fix the geometry every problem shares. Every answer on this page is one of the four intervals hidden in this diagram.

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

Definition Reading the figure above, piece by piece

Look at the horizontal black line — that is the real number line, all possible values of $x$ . Now trace three things on it:

The black dot in the middle is the ==centre $a$ ==. This is the one point where the series is always guaranteed to converge (plug $x = a$ and only the $c_{0}$ term survives).
The red span stretching left and right of the dot is where the series converges. Its half-width — the distance from the dot to either end — is the ==radius $R$ ==. That is what the red double-arrow labelled $R$ measures.
The two hollow circles at $x = a - R$ and $x = a + R$ are the endpoints. They are drawn hollow because we don't yet know whether they belong to the interval — at these two points the tests go silent, so we must check them one at a time by hand. Beyond them (the plain black stretches labelled "diverges") the series blows up.

Every exercise below is a hunt for the black dot ( $a$ ), the red half-width ( $R$ ), and then a case-by-case verdict on the two hollow circles.

Level 1 — Recognition

Recall Solution L1·(a)

WHAT we look at: the shape $(x - a)^{n}$ . The number subtracted from $x$ is $a = 7$ , the centre. WHY it always converges there: plug $x = 7$ . Then $(x - 7)^{n} = 0^{n} = 0$ for every $n \geq 1$ , so only the $n = 0$ term $c_{0}$ survives. A single number is a finite sum, which trivially converges. Answer: centre $a = 7$ ; guaranteed convergence at $x = 7$ (the black dot in the opening figure).

Recall Solution L1·(b)

WHY this formula applies: the Ratio Test version $R = lim_{n \to \infty} ∣ c_{n} / c_{n + 1} ∣$ is valid exactly when that limit exists — and here we are handed its value. WHAT we get: $R = 5$ . The centre is $a = 0$ (no shift in $x^{n}$ ), so the open interval is $(0 - 5, 0 + 5) = (- 5, 5)$ . Answer: $R = 5$ , open interval $(- 5, 5)$ . (Endpoints unknown until we know the $c_{n}$ .)

Level 2 — Application

Recall Solution L2·(a)

Step 1 — WHAT (ratio): with $a_{n} = \frac{( x - 4 ) ^{n}}{n 5 ^{n}}$ , $\frac{a _{n + 1}}{a _{n}} = \frac{∣ x - 4 ∣ ^{n + 1} / (( n + 1 ) 5 ^{n + 1} )}{∣ x - 4 ∣ ^{n} / ( n 5 ^{n} )} = \frac{∣ x - 4∣}{5} \cdot \frac{n}{n + 1} .$ WHY this step: the $∣ x - 4 ∣^{n}$ cancels leaving one factor, the $5^{n}$ leaves one factor of $5$ , and $\frac{n}{n + 1} \to 1$ . Step 2 — from the test's conclusion to $R$ : the limit is $L = \frac{∣ x - 4∣}{5}$ . The Ratio Test says the series converges whenever $L < 1$ and diverges whenever $L > 1$ . So the exact set of $x$ that converge is found by solving the inequality $L < 1$ for the distance $∣ x - 4∣$ : $\frac{∣ x - 4∣}{5} < 1 ⟺ ∣ x - 4∣ < 5.$ That last line is the defining condition $∣ x - a ∣ < R$ of the radius, so we simply read off $R = 5$ . This is why solving $L < 1$ hands us $R$ : the boundary $L = 1$ is exactly the boundary $∣ x - a ∣ = R$ . Centre $a = 4$ , open interval $(- 1, 9)$ . Step 3 — endpoints (substitute!):

$x = 9$ : $\sum \frac{5 ^{n}}{n 5 ^{n}} = \sum \frac{1}{n}$ — the harmonic series, a p-series with $p = 1$ : diverges.
$x = - 1$ : $\sum \frac{( - 5 ) ^{n}}{n 5 ^{n}} = \sum \frac{( - 1 ) ^{n}}{n}$ — by the Alternating Series Test the terms $1/ n$ decrease to $0$ : converges. Answer: $R = 5$ , interval $[- 1, 9)$ .

Recall Solution L2·(b)

Ratio: $\frac{( 2 n + 2 )! x ^{n + 1}}{( 2 n )! x ^{n}} = (2 n + 2) (2 n + 1) ∣ x ∣.$ WHY: $\frac{( 2 n + 2 )!}{( 2 n )!} = (2 n + 2) (2 n + 1)$ , which $\to \infty$ . For any $x \neq = 0$ this limit is $\infty > 1$ , so the series diverges except at $x = 0$ . Answer: $R = 0$ ; converges only at $x = 0$ .

Level 3 — Analysis

Recall Solution L3·(a)

Ratio: $\frac{a _{n + 1}}{a _{n}} = ∣ x - 1∣ \cdot \frac{n}{n + 1} .$ WHY $\frac{n}{n + 1} \to 1$ : inside the root, $\frac{n}{n + 1} = \frac{1}{1 + 1/ n}$ , and as $n \to \infty$ the $1/ n$ shrinks to $0$ , so the fraction $\to 1$ . The square root of something tending to $1$ also tends to $1$ (the root function is continuous at $1$ ). Hence the ratio $\to ∣ x - 1∣$ . From $L < 1$ to $R$ : solving $∣ x - 1∣ < 1$ gives $R = 1$ , centre $a = 1$ , open interval $(0, 2)$ . Endpoint $x = 2$ : $(x - 1) = 1$ , giving $\sum \frac{( - 1 ) ^{n}}{n}$ . Terms $1/ n$ decrease to $0$ ⇒ Alternating Series Test gives convergence. (Note: it is not absolutely convergent, since $\sum 1/ n$ is a $p$ -series with $p = \frac{1}{2} < 1$ : this is conditional.) Endpoint $x = 0$ : $(x - 1) = - 1$ , so $(- 1)^{n} (x - 1)^{n} = (- 1)^{n} (- 1)^{n} = 1$ . The series is $\sum \frac{1}{n}$ — a p-series with $p = \frac{1}{2} < 1$ : diverges. Answer: $R = 1$ , interval $(0, 2]$ .

Recall Solution L3·(b)

Ratio: $\frac{a _{n + 1}}{a _{n}} = \frac{∣ x + 2∣}{3} \cdot \frac{n ^{2}}{( n + 1 ) ^{2}} .$ WHY $\frac{n ^{2}}{( n + 1 ) ^{2}} \to 1$ : write it as $(\frac{n}{n + 1})^{2} = (\frac{1}{1 + 1/ n})^{2}$ . Since $1/ n \to 0$ , the inside $\to 1$ , and squaring $1$ still gives $1$ . Hence the ratio $\to \frac{∣ x + 2∣}{3}$ . From $L < 1$ to $R$ : converge when $\frac{∣ x + 2∣}{3} < 1 ⟺ ∣ x + 2∣ < 3$ : $R = 3$ , centre $a = - 2$ , open interval $(- 5, 1)$ . Endpoint $x = 1$ : $(x + 2) = 3$ , giving $\sum \frac{3 ^{n}}{n ^{2} 3 ^{n}} = \sum \frac{1}{n ^{2}}$ — p-series $p = 2 > 1$ : converges. Endpoint $x = - 5$ : $(x + 2) = - 3$ , giving $\sum \frac{( - 3 ) ^{n}}{n ^{2} 3 ^{n}} = \sum \frac{( - 1 ) ^{n}}{n ^{2}}$ — converges absolutely: converges. Answer: $R = 3$ , interval $[- 5, 1]$ .

Level 4 — Synthesis

Recall Solution L4·(a)

WHAT is subtle: only even powers appear, so the coefficient of $x^{2 n + 1}$ is $0$ . Naively writing $c_{n} / c_{n + 1}$ mixes zero and nonzero terms — dangerous. Fix: apply the Ratio Test directly to the actual terms, treating the whole $x^{2 n}$ as the object. $\frac{x ^{2 (n + 1)} / 4 ^{n + 1}}{x ^{2 n} / 4 ^{n}} = \frac{∣ x ∣ ^{2}}{4} .$ WHY: $x^{2 n + 2} / x^{2 n} = x^{2}$ and $4^{n + 1} / 4^{n} = 4$ . Here the limit $L = \frac{∣ x ∣ ^{2}}{4}$ has no $n$ left in it at all. Solving the test's convergence condition $L < 1$ : $\frac{∣ x ∣ ^{2}}{4} < 1 ⟺ ∣ x ∣^{2} < 4 ⟺ ∣ x ∣ < 2$ . So $R = 2$ , centre $0$ , open interval $(- 2, 2)$ . Sanity check via Geometric Series: the series is $\sum (x^{2} /4)^{n}$ , geometric with ratio $r = x^{2} /4$ ; it converges exactly when $∣ r ∣ < 1$ , i.e. $∣ x^{2} /4∣ < 1$ , i.e. $∣ x ∣ < 2$ — matches. Endpoints $x = \pm 2$ : ratio $r = 4/4 = 1$ , so each is $\sum 1 = 1 + 1 + 1 + \dots$ — the terms don't go to $0$ : diverges (divergence test). Answer: $R = 2$ , interval $(- 2, 2)$ .

Recall Solution L4·(b)

Step 1 — differentiate term by term: $\frac{d}{d x} \frac{x ^{n}}{n} = \frac{n x ^{n - 1}}{n} = x^{n - 1} .$ So $f^{'} (x) = \sum_{n = 1}^{\infty} x^{n - 1} = 1 + x + x^{2} + \dots = \sum_{k = 0}^{\infty} x^{k} .$ Step 2 — WHY the radius is unchanged: this is the term-by-term differentiation theorem from Term-by-term Differentiation and Integration. Its proof rests on the Root Test master formula $\frac{1}{R} = lim sup_{n} n ∣ c_{n} ∣$ . Differentiating turns the coefficient $c_{n}$ into $n c_{n}$ (roughly), so the new radius uses $lim sup_{n} n n ∣ c_{n} ∣$ . But $n n \to 1$ (the $n$ -th root of $n$ tends to $1$ , since $n$ grows far slower than any exponential $r^{n}$ ), so the extra factor of $n$ disappears in the $lim sup$ — leaving $\frac{1}{R}$ exactly as before. That is precisely why differentiation preserves $R$ . Here $R = 1$ still, and indeed this $f^{'}$ is a Geometric Series with ratio $x$ : converges for $∣ x ∣ < 1$ . Step 3 — endpoints CAN change:

$x = 1$ : $\sum 1 = 1 + 1 + \dots$ — diverges (was excluded for $f$ too).
$x = - 1$ : $\sum (- 1)^{k} = 1 - 1 + 1 - \dots$ — partial sums oscillate, terms don't $\to 0$ : diverges (was convergent for the original $f$ !). The lesson: $f$ converged at $x = - 1$ but its derivative $f^{'}$ does not. Differentiation keeps $R$ but can lose endpoints. Answer: $f^{'} (x) = \frac{1}{1 - x}$ on $(- 1, 1)$ ; interval $(- 1, 1)$ .

Level 5 — Mastery

Recall Solution L5·(a)

Why the ratio fails: the ratio $∣ c_{n + 1} / c_{n} ∣$ is $\frac{1}{3 ^{n}}$ (even→odd step) or $3^{n + 1}$ (odd→even step). It oscillates between near- $0$ and huge — no limit exists, so $R = lim ∣ c_{n} / c_{n + 1} ∣$ has nothing to converge to. Why the root works: the Root Test uses $n ∣ c_{n} ∣$ , which is $n 3^{n} = 3 (n even), n 1 = 1 (n odd) .$ This sequence just alternates between $3$ and $1$ — its limit superior (the largest value it keeps returning to) is $β = lim sup_{n \to \infty} n ∣ c_{n} ∣ = 3$ . Cauchy–Hadamard: $\frac{1}{R} = β = 3 \Rightarrow R = \frac{1}{3} .$ Why $lim sup$ and not $lim$ : the $lim sup$ always exists in $[0, \infty]$ , so it is the master tool; the ratio limit is only a convenient special case when things settle down.

Recall Solution L5·(b)

WHAT is subtle: only odd powers $x^{2 n + 1}$ appear (the even-power coefficients are $0$ ). As in L4·(a), we sidestep the gap by applying the Ratio Test to the actual terms $a_{n} = \frac{( - 1 ) ^{n} x ^{2 n + 1}}{( 2 n + 1 )!}$ , treating $x^{2 n + 1}$ as the object. Ratio: $\frac{a _{n + 1}}{a _{n}} = \frac{x ^{2 n + 3} / ( 2 n + 3 )!}{x ^{2 n + 1} / ( 2 n + 1 )!} = \frac{∣ x ∣ ^{2}}{( 2 n + 3 ) ( 2 n + 2 )} .$ WHY: $x^{2 n + 3} / x^{2 n + 1} = x^{2}$ , and $\frac{( 2 n + 1 )!}{( 2 n + 3 )!} = \frac{1}{( 2 n + 3 ) ( 2 n + 2 )}$ (the extra two factorial factors sit in the denominator). Take the limit: for every fixed $x$ , the denominator $(2 n + 3) (2 n + 2) \to \infty$ , so the whole ratio $\to 0$ . Since $0 < 1$ no matter what $x$ is, the Ratio Test gives convergence for all real $x$ . Conclusion: $R = \infty$ ; interval $(- \infty, \infty)$ . Because $R = \infty$ there are no finite endpoints to check — the series converges everywhere (as it must, since it equals $sin x$ , defined for all $x$ ).

Recall Solution L5·(c)

Root values: $n ∣ c_{n} ∣$ is $n 2^{n} = 2$ on the perfect squares ( $n = 0, 1, 4, 9, \dots$ ) and $n 5^{n} = 5$ everywhere else. Which value dominates the $lim sup$ ? The perfect squares get rarer and rarer, but between them the value is $5$ infinitely often. The limit superior is the largest value hit infinitely often, so $β = 5$ . Cauchy–Hadamard: $\frac{1}{R} = 5 \Rightarrow R = \frac{1}{5}$ , centred at $a = 2$ ; open interval $(2 - \frac{1}{5}, 2 + \frac{1}{5}) = (\frac{9}{5}, \frac{11}{5})$ .

Recall checkpoint

Recall Quick self-quiz

Centre of $\sum c_{n} (x + 3)^{n}$ ? ::: $x = - 3$ (set base $x + 3 = 0$ ). Given $lim ∣ c_{n} / c_{n + 1} ∣ = 5$ and centre $0$ , the open interval is? ::: $(- 5, 5)$ . At $∣ x - a ∣ = R$ the ratio test gives limit $=$ ? ::: exactly $1$ — inconclusive; test endpoints separately. If the coefficient ratio oscillates, which formula still works? ::: $1/ R = lim sup_{n \to \infty} n ∣ c_{n} ∣$ (Cauchy–Hadamard / Root). Does term-by-term differentiation change $R$ ? ::: No — $R$ is unchanged (because $n n \to 1$ ), but endpoints must be re-tested. $\sum x^{2 n} / 4^{n}$ has radius? ::: $R = 2$ ; geometric with ratio $x^{2} /4$ .