A power series is a "polynomial with infinitely many terms." Once you know where it converges (its interval, from a Ratio Test computation exactly like Week 15's), you can differentiate and integrate it term by term. Taylor series push this further: they build the power series of an already-known function from nothing but its derivatives at one point — turning transcendental functions into computable, truncatable polynomials.
16.1 Power Series
A power series centered at x=a is a series of the form
where the cn are constants. For each fixed x, this is an ordinary numerical series — exactly the kind Week 15's tests apply to. A power series therefore does not converge or diverge absolutely; it converges for some values of x and diverges for others, and mapping out exactly which x is the first task.
16.2 Radius and Interval of Convergence
Every power series has a radius of convergence R (possibly 0 or ∞) such that the series converges absolutely for |x−a| < R and diverges for |x−a| > R. The standard way to find R is the Ratio Test from Week 15 (§15.6), applied to the terms cn(x−a)n. The interval of convergence is this interval together with whichever endpoints (if any) also converge — the endpoints must be checked separately, one at a time, using any test from Week 15, since the Ratio Test itself is silent exactly at ρ=1.
Three possible shapes. The interval of convergence is always one of: a single point x=a (R=0), the entire real line (R=∞), or a finite interval centered at a with radius R, open, half-open, or closed depending on the endpoint check.
16.3 Term-by-Term Differentiation and Integration
Inside its interval of convergence, a power series behaves like a polynomial for calculus purposes: it can be differentiated or integrated term by term (Week 4 & Week 9's rules applied to each term individually), and the resulting series has the same radius of convergence R (though endpoint behavior can change). This is what makes power series usable as actual functions rather than mere formal sums.
16.4 Taylor and Maclaurin Series
If f has derivatives of every order at x=a, its Taylor series generated at a is
Every coefficient comes directly from an ordinary derivative evaluated at the single point a (Week 4's differentiation rules). A Maclaurin series is simply the special case a=0. The n-th Taylor polynomial Pn(x) is the partial sum through the n-th term — a genuine (finite) polynomial that approximates f near a, improving as n grows.
16.5 Common Maclaurin Series
Four series, derived directly from Eq. 16.2 by repeated differentiation, recur constantly and are worth memorizing:
Eq. 16.3d is just the geometric series from Week 14 (§14.6) read as a power series in x; its radius of convergence, R=1, was already known from that section without invoking the Ratio Test at all.
16.6 Convergence of Taylor Series and the Remainder
A Taylor series generated by f is not guaranteed to converge to f(x) — it may converge to a different value, or diverge, outside a certain range. Writing f(x) = Pn(x) + Rn(x), the series converges to f(x) at a given x exactly when the remainder Rn(x) → 0 as n → ∞. Within the disk of convergence, all four series in §16.5 do converge to the named function — this is why they can be used for exact evaluation, not just approximation.
16.7 Evaluating Non-Elementary Integrals with Series
Week 13 (§13.4–13.7) built a toolbox of integration techniques, but some elementary functions — such as e−x2, sin(x2), or (sin x)/x — simply have no antiderivative expressible in elementary terms. No technique from Week 13 can close these integrals in finite form. Substituting the Maclaurin series (§16.5) for the integrand and integrating the resulting power series term by term (§16.3), however, always works: each term is an elementary power of x, and the result is a convergent series representing the exact value of the definite integral, computable to any desired precision by truncation.
16.8 Worked Examples
Example 16.1 — Radius and Interval of Convergence
Find the radius and interval of convergence of ∑n=1∞ (x−2)n/(n·3n).
Step 1 — Apply the Ratio Test (Week 15, §15.6) to the general term
limn→∞ |x−2|n+1/((n+1)3n+1)|x−2|n/(n·3n) = limn→∞ n|x−2|3(n+1) = |x−2|3
Step 2 — Require ρ < 1
|x−2|/3 < 1 ⇔ |x−2| < 3, so R=3 and the series converges absolutely for x ∈ (−1, 5).
Step 3 — Check both endpoints separately
At x=−1: the series becomes ∑(−1)â¿/n, the alternating harmonic series, which converges (Week 15, §15.8). At x=5: the series becomes ∑1/n, the harmonic series, which diverges (Week 15, §15.2).
Step 4 — Conclude
Radius of convergence R=3; interval of convergence [−1, 5).
Example 16.2 — Building a Maclaurin Series from Derivatives
Find the Maclaurin series for f(x) = ln(1+x).
Step 1 — Recognize f′(x) as a geometric series
f′(x) = 1/(1+x) = 1/(1−(−x)) = ∑n=0∞ (−x)n, for |x|<1 (Eq. 16.3d with x replaced by −x).
Step 2 — Integrate term by term (§16.3)
ln(1+x) = ∫0x f′(t) dt = ∑n=0∞ (−1)n xn+1n+1
Step 3 — State the result
ln(1+x) = x − x2/2 + x3/3 − x4/4 + …, valid for −1 < x ≤ 1 (found without computing a single higher derivative).
Example 16.3 — A Non-Elementary Integral via Series (§16.7)
Estimate ∫01 e−x2 dx to three decimal places. No elementary antiderivative exists for this integrand, so none of Week 13's techniques apply directly.
Step 1 — Substitute into the Maclaurin series for e^x (Eq. 16.3a)
e−x2 = ∑n=0∞ (−x2)nn! = 1 − x2 + x42! − x63! + …
Step 2 — Integrate term by term from 0 to 1
∫01 e−x2 dx = 1 − 1/3 + 1/(2!·5) − 1/(3!·7) + 1/(4!·9) − … = 1 − 0.3333 + 0.1000 − 0.0238 + 0.0044 − …
Step 3 — Apply the Alternating Series estimate (Week 15, §15.8)
For an alternating series meeting the Alternating Series Test, truncation error is bounded by the first omitted term. Stopping after the 1/(4!·9) term (≈0.0044) keeps the error below the next term, ≈0.0007, comfortably within three-decimal accuracy.
Step 4 — Sum
∫₀¹ e^(−x²) dx ≈ 0.747, exact to three decimal places — obtained with no antiderivative at all, only term-by-term series integration.
16.9 Nomenclature
| Symbol / Term | Meaning |
|---|---|
| Power series | ∑c_n(x−a)^n |
| Radius of convergence R | Half-width of the interval on which the series converges absolutely |
| Interval of convergence | Radius interval plus any convergent endpoints |
| Taylor series | ∑[f^(n)(a)/n!](x−a)^n |
| Maclaurin series | Taylor series with a=0 |
| P_n(x) | n-th Taylor polynomial (partial sum of the Taylor series) |
| R_n(x) | Remainder, f(x) − P_n(x) |
16.10 Problems
- Radius of convergence. Find the radius and interval of convergence of ∑x^n/n².
- Radius of convergence. Find the radius and interval of convergence of ∑n!x^n.
- Endpoint check. Find the interval of convergence of ∑(x+1)^n/√n, checking both endpoints.
- Maclaurin from Eq. 16.3a. Use Eq. 16.3a to write the Maclaurin series for e^(2x).
- Maclaurin from Eq. 16.3b. Use Eq. 16.3b to write the Maclaurin series for sin(x²).
- Building from derivatives. Find the first four nonzero terms of the Taylor series for cos x centered at a=π.
- Geometric series as power series. Use Eq. 16.3d to find a power series for 1/(1+x²), and state its radius of convergence.
- Term-by-term integration. Integrate your answer to Problem 7 to find a power series for arctan x.
- Non-elementary integral. Use a Maclaurin series to estimate ∫₀¹ sin(x²) dx to three decimal places.
- Non-elementary integral. Use a Maclaurin series to estimate ∫₀·¹ (sin x)/x dx to three decimal places (note the integrand's removable behavior at x=0).
16.11 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. §9.7 (Power Series), §9.8 (Taylor and Maclaurin Series), §9.9 (Convergence of Taylor Series), §9.10 (Applications of Taylor Series — evaluating non-elementary integrals only, per this term's schedule)
- Pearson MyLab Math — Week 16 homework assigned directly from the topics above. See the course calendar for exact due dates.