Series & Sequences · Convergence

Power Series

Infinite polynomials that represent functions, each with its own radius of convergence, interval of convergence, and a complete toolkit for differentiation and integration.

Radius of Convergence Interval of Convergence Ratio Test Method Term-by-Term Differentiation Term-by-Term Integration
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  1. What Is a Power Series?
  2. Radius of Convergence
  3. Finding the Radius via the Ratio Test
  4. The Interval of Convergence, Endpoints
  5. Interactive Convergence Visualiser
  6. Functions Represented by Power Series
  7. Differentiation of Power Series
  8. Integration of Power Series
  9. Worked Examples
  10. Common Mistakes
  11. 10-Question Quiz
  12. Continue in Series & Sequences

§ 01What Is a Power Series?

A power series is an infinite polynomial: a sum of terms of the form cₙ(x − a)ⁿ. Unlike a finite polynomial, it may only converge for certain values of x, making the domain itself an object of study.

Every function you have encountered in calculus, eˣ, sin x, cos x, ln(1+x), can be expressed as a power series around an appropriate centre. Power series are the bridge between the world of sequences and the world of functions, and they underpin everything from Taylor approximations to the solutions of differential equations.

General Power Series Centred at a
Σn=0∞ cn(x − a)n = c0 + c1(x−a) + c2(x−a)² + c3(x−a)³ + ···

Here a is the centre of the series, and {cn} is the sequence of coefficients. When a = 0, the series is a Maclaurin series (or simply a "power series in x"). For general a, it is a Taylor series centred at a.

The key question for any power series is: for which values of x does the series converge? The answer is always an interval (possibly a single point, possibly all of ℝ) centred at a, the interval of convergence.

Power series behave like polynomials inside their interval of convergence Within the open interval (a−R, a+R), a power series is infinitely differentiable and you can differentiate or integrate it term by term, just as you would a polynomial. This is one of the most powerful tools in analysis.

§ 02Radius of Convergence

For every power series Σ cₙ(x−a)ⁿ, there exists a number R ∈ [0, ∞] called the radius of convergence such that:

Radius of Convergence — Three Cases
|x − a| < R  ⟹  the series converges absolutely |x − a| > R  ⟹  the series diverges |x − a| = R  ⟹  no conclusion — must test each endpoint separately

The three special cases for R are:

  • R = 0: the series converges only at x = a (a single point).
  • R = ∞: the series converges for all x ∈ ℝ (the entire number line).
  • 0 < R < ∞: the series converges on the open interval (a−R, a+R) and diverges outside it.

The radius of convergence is a property of the coefficients {cₙ} alone: it does not depend on what happens at the endpoints. This is why it is called the radius: it describes the half-width of the interval of convergence before endpoint behaviour is decided.

Hadamard's Formula (theoretical)
R = 1 / limsupn→∞ |cn|1/n

Hadamard's formula gives the exact radius, but it requires computing a limsup, which can be difficult. In practice, we use the Ratio Test or Root Test on the series directly: this is almost always more tractable.

§ 03Finding the Radius via the Ratio Test

The most common method. Apply the Ratio Test to the power series treated as a series in x:

Ratio Test for Power Series
L = limn→∞ |cn+1(x−a)n+1| / |cn(x−a)n| = |x−a| · limn→∞ |cn+1/cn| Converges absolutely when L < 1, i.e. |x−a| < lim |cn/cn+1| ∴   R = limn→∞ |cn / cn+1|

Step-by-step procedure

  • Step 1. Write out the ratio |an+1/an| where an = cₙ(x−a)ⁿ.
  • Step 2. Factor out |x−a| from the ratio.
  • Step 3. Compute the limit of the remaining coefficient ratio.
  • Step 4. Set the result < 1 and solve for |x−a| to find R.
  • Step 5. Test the two endpoints x = a ± R separately.
Root Test alternative When coefficients involve nth powers (like (2/3)ⁿ), the Root Test is cleaner: R = 1/lim|cₙ|^(1/n). Use it whenever the coefficients are explicitly raised to the nth power.

§ 04The Interval of Convergence — Checking Endpoints

Once R is found, the interval of convergence is one of four possibilities depending on what happens at x = a − R and x = a + R:

Left endpoint x = a−RRight endpoint x = a+RInterval of Convergence
DivergesDiverges(a−R, a+R)
ConvergesDiverges[a−R, a+R)
DivergesConverges(a−R, a+R]
ConvergesConverges[a−R, a+R]

To test an endpoint x = a ± R, substitute that x-value into the series. The result is a constant-term series (x is now a specific number), which you test with any standard test, p-series, alternating series test, divergence test, etc.

The Ratio Test is inconclusive at endpoints When |x−a| = R exactly, the Ratio Test gives L = 1, which tells you nothing. You must use a different test at each endpoint individually. This step is the most commonly forgotten in exam questions.

§ 05Interactive Convergence Visualiser

The chart below plots the partial sum approximation SN(x) of the power series against the true function f(x). Use the slider to increase N and observe how the approximation improves inside the radius of convergence: and breaks down outside it.

Interactive · Power Series Partial Sum Approximation

Partial sums of selected power series vs. true function

Series
eˣ
Radius R
∞
N terms used
6
Centre a
0

Notice that for series with finite R = 1 (the geometric series and ln(1+x)), the approximation visibly diverges for |x| > 1, no matter how many terms you use. For R = ∞ (eˣ, sin x, cos x), the partial sums eventually converge everywhere, though you need more terms further from the centre.

§ 06Functions Represented by Power Series

The following power series centred at 0 (Maclaurin series) are the most important in all of calculus. Memorising them unlocks fast computation of limits, integrals, and differential equations.

FunctionPower SeriesRadius R
eˣΣ xⁿ/n! = 1 + x + x²/2! + x³/3! + ···∞
sin xΣ (−1)ⁿx²ⁿ⁺¹/(2n+1)! = x − x³/6 + x⁵/120 − ···∞
cos xΣ (−1)ⁿx²ⁿ/(2n)! = 1 − x²/2 + x⁴/24 − ···∞
1/(1−x)Σ xⁿ = 1 + x + x² + x³ + ···1
ln(1+x)Σ (−1)ⁿ⁺¹xⁿ/n = x − x²/2 + x³/3 − ···1
arctan xΣ (−1)ⁿx²ⁿ⁺¹/(2n+1) = x − x³/3 + x⁵/5 − ···1
(1+x)ᵏΣ C(k,n)xⁿ = 1 + kx + k(k−1)x²/2! + ···1 (k∉ℕ)

These series can be combined, composed, multiplied, and transformed to produce power series for more complex functions. The key operations (differentiation and integration) are covered in the next two sections.

§ 07Differentiation of Power Series

Inside the open interval of convergence (a−R, a+R), a power series can be differentiated term by term, and the resulting series has the same radius of convergence R:

Term-by-Term Differentiation
If f(x) = Σn=0∞ cn(x−a)n, then: f'(x) = Σn=1∞ n·cn(x−a)n−1

Note the sum starts at n = 1 because the n = 0 term (the constant c₀) differentiates to zero. The differentiated series has the same radius R, though the interval of convergence may change at endpoints (endpoints that converged may now diverge, or vice versa).

This means that if f(x) = Σ cₙ(x−a)ⁿ within its interval, then f is infinitely differentiable there, and every coefficient is determined by the derivatives: cₙ = f⁽ⁿ⁾(a)/n!, which is precisely the Taylor series formula.

§ 08Integration of Power Series

Similarly, a power series can be integrated term by term inside its open interval of convergence. The resulting series also has radius R:

Term-by-Term Integration
∫ f(x) dx = Σn=0∞ cn(x−a)n+1/(n+1) + C

This is extraordinarily useful for integrating functions that have no elementary antiderivative. For example, ∫ sin(x²) dx, ∫ e−x² dx, and ∫ sin(x)/x dx all lack closed forms, but their power series can be integrated term by term to produce a series representation of the antiderivative.

Application: evaluating definite integrals By integrating a power series term by term and applying definite limits, we can compute integrals like ∫₀¹ sin(x²)dx or ∫₀0.5 e−x²dx to any desired accuracy — as long as we stay within the interval of convergence.
✦   ✦   ✦

§ 09Worked Examples

Twelve worked examples covering radius and interval of convergence, endpoint testing, and operations on power series.

Example 01 — Radius by Ratio Test (polynomial coefficients)
Find the radius and interval of convergence of Σn=0∞ xⁿ/n.
  1. Apply the Ratio Test: |an+1/an| = |xn+1/(n+1)| / |xⁿ/n| = |x|·n/(n+1).
  2. Limit: lim n/(n+1) = 1, so L = |x|. Converges when |x| < 1. ∴ R = 1.
  3. Check x = 1: Σ 1/n, diverges (harmonic series).
  4. Check x = −1: Σ (−1)ⁿ/n, converges by AST.
R = 1,   Interval: [−1, 1)
Example 02 — Radius by Ratio Test (factorial coefficients)
Find R for Σn=0∞ xⁿ/n!
  1. Ratio: |xn+1/(n+1)!| / |xⁿ/n!| = |x|/(n+1).
  2. Limit: lim |x|/(n+1) = 0 for all x.
  3. L = 0 < 1 for all x ∈ ℝ. ∴ R = ∞.
  4. Interval of convergence: (−∞, ∞): this is the Maclaurin series for eˣ.
R = ∞,   Interval: (−∞, ∞)
Example 03 — Series centred at a ≠ 0
Find R and the interval of convergence for Σn=0∞ (x − 3)ⁿ / 4ⁿ.
  1. Ratio: |(x−3)n+1/4n+1| / |(x−3)ⁿ/4ⁿ| = |x−3|/4.
  2. Converges when |x−3|/4 < 1, i.e. |x−3| < 4. ∴ R = 4, centre a = 3.
  3. Check x = 7 (x−3 = 4): Σ 1, diverges.
  4. Check x = −1 (x−3 = −4): Σ (−1)ⁿ, diverges (terms don't → 0).
R = 4,   Interval: (−1, 7)
Example 04 — Coefficient involves n²
Find R for Σn=1∞ n²xⁿ.
  1. Ratio: |(n+1)²xn+1| / |n²xⁿ| = |x| · (n+1)²/n².
  2. lim (n+1)²/n² = 1. So L = |x|.
  3. Converges when |x| < 1. ∴ R = 1.
  4. Endpoints: at x = ±1, terms n²(±1)ⁿ → ∞, so both endpoints diverge.
R = 1,   Interval: (−1, 1)
Example 05 — Power of n in the coefficient
Find R for Σn=0∞ (2x)ⁿ = Σ 2ⁿxⁿ using the Root Test.
  1. Root Test: lim |2ⁿxⁿ|1/n = lim 2|x| = 2|x|.
  2. Converges when 2|x| < 1, i.e. |x| < 1/2. ∴ R = 1/2.
  3. This is the geometric series 1/(1−2x), valid for |x| < 1/2.
R = 1/2,   Interval: (−1/2, 1/2)
Example 06 — Deriving 1/(1−x)² by differentiation
Using the known series 1/(1−x) = Σxⁿ, find the power series for 1/(1−x)².
  1. Note d/dx[1/(1−x)] = 1/(1−x)².
  2. Differentiate Σxⁿ term by term: d/dx[Σ xⁿ] = Σ nxⁿ⁻¹ = Σn=1∞ nxⁿ⁻¹.
  3. Re-index: Σn=0∞ (n+1)xⁿ.
1/(1−x)² = Σn=0∞(n+1)xⁿ,   |x| < 1
Example 07 — Deriving ln(1+x) by integration
Using 1/(1+x) = Σ(−x)ⁿ = Σ(−1)ⁿxⁿ, derive the power series for ln(1+x).
  1. Since d/dx[ln(1+x)] = 1/(1+x) = Σ(−1)ⁿxⁿ, integrate both sides:
  2. ln(1+x) = ∫Σ(−1)ⁿxⁿ dx = Σ(−1)ⁿ xn+1/(n+1) + C
  3. At x = 0: ln(1) = 0 = 0 + C, so C = 0.
  4. ln(1+x) = Σn=0∞(−1)ⁿxn+1/(n+1) = x − x²/2 + x³/3 − ···
ln(1+x) = Σn=1∞(−1)n+1xⁿ/n,   −1 < x ≤ 1
Example 08 — Power series for arctan x by integration
Use 1/(1+x²) = Σ(−1)ⁿx²ⁿ to find the series for arctan x.
  1. d/dx[arctan x] = 1/(1+x²) = Σ(−1)ⁿx²ⁿ for |x| < 1.
  2. Integrate: arctan x = Σ(−1)ⁿ x2n+1/(2n+1) + C.
  3. At x = 0: arctan(0) = 0 = C. So C = 0.
  4. arctan x = x − x³/3 + x⁵/5 − x⁷/7 + ···, valid for −1 ≤ x ≤ 1.
  5. At x = 1: arctan 1 = π/4 = 1 − 1/3 + 1/5 − ···: Leibniz's formula!
arctan x = Σn=0∞(−1)ⁿx2n+1/(2n+1),   |x| ≤ 1
Example 09 — Using a power series to evaluate an integral
Express ∫₀1 sin(x²) dx as a series and find its value to 4 decimal places.
  1. sin u = Σ(−1)ⁿu2n+1/(2n+1)!. Substitute u = x²:
  2. sin(x²) = Σ(−1)ⁿx4n+2/(2n+1)!
  3. Integrate term by term from 0 to 1:
    ∫₀¹ sin(x²)dx = Σ(−1)ⁿ/(2n+1)! · 1/(4n+3)
  4. Terms: 1/3 − 1/(7·6) + 1/(11·120) − ···
  5. = 0.33333 − 0.02381 + 0.00076 − 0.00001 + ··· ≈ 0.3103
∫₀¹ sin(x²) dx ≈ 0.3103
Example 10 — R = 0 example
Show that Σ n! xⁿ has radius of convergence R = 0.
  1. Ratio: |(n+1)!xn+1/(n!xⁿ)| = (n+1)|x|.
  2. lim(n+1)|x| = ∞ for any x ≠ 0. So L = ∞ > 1 for all x ≠ 0.
  3. Series diverges for every x ≠ 0. At x = 0, the series trivially = 1.
R = 0, converges only at x = 0
Example 11 — Finding coefficients from a known function
Find the power series for f(x) = x/(1−x²) centred at 0.
  1. Write x/(1−x²) = x · 1/(1−x²).
  2. Use 1/(1−u) = Σuⁿ with u = x²: 1/(1−x²) = Σ x2n.
  3. Multiply by x: x/(1−x²) = Σ x2n+1 = x + x³ + x⁵ + x⁷ + ···
  4. Valid for |x²| < 1, i.e. |x| < 1.
x/(1−x²) = Σn=0∞ x2n+1,   |x| < 1
Example 12 — Endpoint analysis with p-series
Find the interval of convergence of Σn=1∞ xⁿ / n3/2.
  1. Ratio: |xn+1/(n+1)3/2| / |xⁿ/n3/2| = |x|·(n/(n+1))3/2 → |x|.
  2. R = 1.
  3. x = 1: Σ 1/n3/2: p-series with p = 3/2 > 1. Converges. ✓
  4. x = −1: Σ (−1)ⁿ/n3/2, alternating, bn = 1/n3/2 → 0, decreasing. Converges by AST. ✓
R = 1,   Interval: [−1, 1]

§ 10Common Mistakes

Mistake 1 — Forgetting to test endpoints The Ratio Test gives an open interval. The endpoints x = a ± R are never determined by the Ratio Test — they require separate tests. Missing this costs marks every time.
Mistake 2 — Confusing radius with interval R = 3 centred at a = 2 gives the interval (−1, 5), not (−3, 3). Always write the interval as (a−R, a+R) with the actual centre substituted.
Mistake 3 — Differentiating incorrectly d/dx[Σcₙxⁿ] = Σn·cₙxn−1, starting from n = 1 (not n = 0). The n = 0 term is a constant and differentiates to zero.
Mistake 4 — Applying operations outside the interval Term-by-term differentiation and integration only work inside the open interval of convergence. Results at endpoints must be verified separately.

§ 1110-Question Quiz

Select one answer per question then click Check All Answers.

Score: 0 / 10
Question 01
The radius of convergence of Σ xⁿ/n! is:
Question 02
For Σ nxⁿ, what is R?
Question 03
Σ (x−2)ⁿ / 3ⁿ has centre a = 2 and R = 3. What is the open interval of convergence?
Question 04
When is the Ratio Test inconclusive for a power series?
Question 05
The power series for 1/(1−x) = Σxⁿ. What is the series for d/dx[1/(1−x)] = 1/(1−x)²?
Question 06
If f(x) = Σ cₙxⁿ with R = 5, what is the radius of convergence of f'(x)?
Question 07
The power series Σ n! xⁿ has radius of convergence:
Question 08
arctan x = Σ (−1)ⁿx²ⁿ⁺¹/(2n+1). At x = 1 this gives the famous formula:
Question 09
To find a power series for e−x², which substitution works in the series for eˣ?
Question 10
Σ xⁿ/n converges at x = −1 but diverges at x = 1. Its interval of convergence is:
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§ 12Continue in Series & Sequences

You are herePower SeriesRadius & interval of convergence, operations Next →Maclaurin SeriesPower series at a = 0 for standard functions SeriesTaylor SeriesPower series centred at any point a ← PrerequisiteAlternating SeriesNeeded for endpoint tests of ln(1+x) etc. SeriesRatio & Root TestsThe primary tools for finding R SeriesGeometric SeriesSource of the 1/(1−x) power series Seriesp-Series TestUsed for endpoint convergence checks DEs →Differential EquationsPower series solutions to ODEs

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