Sequences, Series, Power Series, Taylor's Theorem and the Mean Value Theorems

This chapter belongs to the Engineering Sciences (XE) paper and to Section XE0 (Engineering Mathematics), the 15-mark part common to every candidate whichever two optional sections are chosen. It answers the parts of XE0.2 that the shared mathematics chapters do not teach: “Mean value theorems; Taylor's theorem; Fundamental theorem and mean value theorem of integral calculus; … Convergence of sequences and series; Tests of convergence of series with non-negative terms (Cauchy Condensation Test, Ratio, Root and integral tests); Power series; Taylor's series.” Limits, L'Hospital's rule, Rolle and Lagrange theorems and maxima and minima are already taught in the shared single-variable chapters. Here you learn convergence of sequences and series and the four tests, the radius and interval of a power series, Taylor's theorem with the Lagrange remainder, Cauchy's mean value theorem and the two theorems of integral calculus. The numericals are sums of series, radius of convergence, the number of Taylor terms needed for a stated error, a Taylor estimate of a square root and the point promised by Cauchy's theorem.

1. Convergence of sequences and series

A sequence (aₙ) converges to L if aₙ gets and stays arbitrarily close to L as n → ∞; otherwise it diverges. A monotone sequence that is bounded always converges (the monotone convergence theorem). Standard limits: rⁿ → 0 for |r| < 1, n^(1/n) → 1, (1 + 1/n)ⁿ → e, and xⁿ/n! → 0 for every x.

A series Σaₙ converges if its partial sums Sₙ = a₁ + … + aₙ converge, and the limit is its sum. The geometric series Σn≥0 rⁿ = 1/(1 − r) for |r| < 1 and diverges otherwise. The p-series Σ1/nᵖ converges if and only if p > 1. Σ n/3ⁿ = (1/3)/(1 − 1/3)² = 3/4, from Σ n xⁿ = x/(1 − x)² for |x| < 1.

⚠️ aₙ → 0 is necessary, not sufficient
If aₙ does not tend to 0 the series diverges. If aₙ → 0 nothing is decided: Σ1/n has terms → 0 yet diverges (its partial sum Hₙ grows like ln n; H₁₂₃₆₇ is only about 10).

Absolute and conditional convergence. Σaₙ is absolutely convergent if Σ|aₙ| converges, and absolute convergence implies convergence. The alternating series test: if bₙ decreases to 0 then Σ(−1)ⁿbₙ converges. So Σ(−1)ⁿ/n converges (to −ln 2) but only conditionally, since Σ1/n diverges.

2. Tests for series with non-negative terms

For aₙ ≥ 0 the partial sums only increase, so the series either converges or diverges to +∞, and the tests below decide which. Comparison: if 0 ≤ aₙ ≤ bₙ and Σbₙ converges then Σaₙ converges; if Σaₙ diverges so does Σbₙ. Limit comparison: if aₙ/bₙ → c with 0 < c < ∞ the two series behave alike.

The four tests named in the syllabus
TestStatementVerdict
RatioL = lim aₙ₊₁/aₙL < 1 converges; L > 1 diverges; L = 1 no decision
RootL = lim aₙ^(1/n)Same three cases as the ratio test
Integralaₙ = f(n), f positive, continuous, decreasing for x ≥ 1Σaₙ and ∫₁^∞ f dx converge or diverge together
Cauchy condensationaₙ non-negative and decreasingΣaₙ and Σ2ᵏ a2ᵏ converge or diverge together

Worked examples. Σn²/3ⁿ: aₙ₊₁/aₙ = (n + 1)²/(3n²) → 1/3 < 1, converges. Σn!/nⁿ: aₙ₊₁/aₙ = (1 + 1/n)^(−n) → 1/e < 1, converges. Σ(n/(2n + 1))ⁿ: aₙ^(1/n) = n/(2n + 1) → 1/2 < 1, converges by the root test. Σ1/(n ln n), n ≥ 2: condensation gives 2ᵏ/(2ᵏ·k ln 2) = 1/(k ln 2), a harmonic series, so it diverges; the integral test agrees since ∫dx/(x ln x) = ln(ln x) → ∞. Σ1/(n(ln n)²): condensed terms 1/(k²(ln 2)²), so it converges.

⚠️ L = 1 decides nothing
Σ1/n (diverges) and Σ1/n² (converges) both give ratio and root limit 1. Use the integral or condensation test (or comparison with a p-series) when the ratio test returns 1. The ratio test is also useless for rational terms such as 1/(n² + 1).

3. Power series: radius and interval of convergence

A power series Σaₙ(x − c)ⁿ converges for |x − c| < R, diverges for |x − c| > R, and the endpoints x = c ± R must be tested separately. The radius is R = lim |aₙ/aₙ₊₁| or R = 1/lim |aₙ|^(1/n), whenever the limit exists; R = 0 and R = ∞ are allowed. Inside the interval the series may be differentiated and integrated term by term and the radius does not change.

Example 1. Σ(x − 2)ⁿ/(n·3ⁿ): aₙ = 1/(n3ⁿ), so |aₙ/aₙ₊₁| = 3(n + 1)/n → 3 and R = 3, centre 2, open interval (−1, 5). At x = −1 the series is Σ(−1)ⁿ/n, which converges (alternating test); at x = 5 it is Σ1/n, which diverges. The interval of convergence is [−1, 5).

Example 2. Σx²ⁿ/4ⁿ is geometric with ratio x²/4, so it converges for |x| < 2, R = 2. Do not apply R = lim|aₙ/aₙ₊₁| blindly when alternate coefficients vanish; use the ratio test on the whole term instead. Σxⁿ/n! has R = ∞, Σn!xⁿ has R = 0, and Σ(2n)!xⁿ/(n!)² has aₙ₊₁/aₙ = (2n + 2)(2n + 1)/(n + 1)² → 4, so R = 1/4.

4. Taylor's theorem and Taylor's series

Taylor's theorem (Lagrange form). If f has n + 1 continuous derivatives near a, then f(x) = Σk=0n f⁽ᵏ⁾(a)(x − a)ᵏ/k! + Rₙ(x), with Rₙ(x) = f⁽ⁿ⁺¹⁾(ξ)(x − a)ⁿ⁺¹/(n + 1)! for some ξ between a and x. The polynomial is the best local fit; the remainder gives a guaranteed error bound. If Rₙ → 0 the Taylor series Σf⁽ᵏ⁾(a)(x − a)ᵏ/k! represents f; about a = 0 it is the Maclaurin series.

Maclaurin series to memorise
FunctionSeriesValid for
eˣ1 + x + x²/2! + x³/3! + …all x
sin xx − x³/3! + x⁵/5! − …all x
cos x1 − x²/2! + x⁴/4! − …all x
ln(1 + x)x − x²/2 + x³/3 − …−1 < x ≤ 1
1/(1 − x)1 + x + x² + …|x| < 1
(1 + x)ᵐ1 + mx + m(m − 1)x²/2! + …|x| < 1

Worked numericals. (i) The Taylor polynomial of √x about a = 4 to second order: f = 2, f′ = 1/(2√x) = 1/4, f″ = −1/(4x^(3/2)) = −1/32. So P₂(x) = 2 + (x − 4)/4 − (x − 4)²/64, and P₂(4.1) = 2 + 0.025 − 0.00015625 = 2.02484375 (true value 2.0248457). (ii) How many terms of eˣ at x = 1 guarantee an error below 10⁻⁴? Rₙ(1) ≤ e/(n + 1)! < 3/(n + 1)!. For n = 6, 3/7! = 5.95 × 10⁻⁴, too big; for n = 7, 3/8! = 7.44 × 10⁻⁵, enough. So n = 7. (iii) (cos x − 1 + x²/2)/x⁴: cos x = 1 − x²/2 + x⁴/24 − …, so the limit is 1/24.

🧠 Build series, do not differentiate
To get a Taylor coefficient or a limit, multiply or substitute known series: eˣ sin x = (1 + x + x²/2 + …)(x − x³/6 + …) = x + x² + x³/3 + …, so the x³ coefficient is 1/3 and f‴(0) = 6 × 1/3 = 2. That is faster than differentiating a product three times.

5. Cauchy's mean value theorem

Rolle's theorem (f(a) = f(b) gives f′(c) = 0) and Lagrange's theorem (f′(c) = (f(b) − f(a))/(b − a)) need f continuous on [a, b] and differentiable on (a, b). Cauchy's mean value theorem extends this to two functions: if f, g are continuous on [a, b], differentiable on (a, b) and g′ ≠ 0 on (a, b), then for some c in (a, b), f′(c)/g′(c) = (f(b) − f(a))/(g(b) − g(a)). Taking g(x) = x recovers Lagrange's theorem, and L'Hospital's rule is proved from it.

Example. f = x², g = x³ on [1, 2]: the right side is (4 − 1)/(8 − 1) = 3/7, and f′/g′ = 2c/(3c²) = 2/(3c). Setting 2/(3c) = 3/7 gives c = 14/9 ≈ 1.556, which lies in (1, 2), as the theorem promises. Note that g′ = 3x² ≠ 0 on (1, 2); the hypothesis g′ ≠ 0 matters, and a point where both derivatives vanish breaks the quotient.

6. Fundamental theorem and mean value theorem of integral calculus

Fundamental theorem, part 1. If f is continuous on [a, b], then F(x) = ∫ₐˣ f(t) dt is differentiable and F′(x) = f(x). With variable limits, the chain rule gives d/dx ∫ₐu(x) f(t) dt = f(u(x))·u′(x). Part 2. If G′ = f then ∫ₐᵇ f dx = G(b) − G(a). Example: d/dx ∫₀x² e^(−t²) dt = e^(−x⁴)·2x, and at x = 1 this equals 2/e ≈ 0.7358.

Mean value theorem of integral calculus. If f is continuous on [a, b], there is c in [a, b] with ∫ₐᵇ f dx = f(c)(b − a); f(c) is the average value of f. For f = x² on [0, 3] the integral is 9, the average is 3, and c = √3 ≈ 1.732. Continuity is essential: a function with a jump may never take its own average value.

🎯 Why differentiation and integration undo each other
A thin strip of area under f between x and x + h has area close to f(x)·h, so the rate of growth of the accumulated area is f(x). Part 2 follows because two functions with the same derivative differ by a constant, which cancels in G(b) − G(a).

Key takeaways

  • Σaₙ needs aₙ → 0 but that alone is never enough; Σ1/nᵖ converges iff p > 1, and a geometric series needs |r| < 1.
  • Ratio and root tests give L < 1 convergent, L > 1 divergent, L = 1 undecided; then use the integral test, Cauchy condensation (Σaₙ with Σ2ᵏa2ᵏ) or comparison.
  • A power series has a radius R; test both endpoints separately; term-by-term differentiation keeps R.
  • Taylor: f = Σf⁽ᵏ⁾(a)(x − a)ᵏ/k! + f⁽ⁿ⁺¹⁾(ξ)(x − a)ⁿ⁺¹/(n + 1)!; the remainder gives the error bound. Learn the six Maclaurin series.
  • Cauchy's MVT: f′(c)/g′(c) = Δf/Δg with g′ ≠ 0; it contains Lagrange's theorem (g = x) and underlies L'Hospital's rule.
  • Fundamental theorem: d/dx ∫ₐu(x) f = f(u)u′; the integral mean value is f(c) = average of f.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Which of the following series converges?

    1. Σ 1/√n for n ≥ 1
    2. Σ 1/n^1.5 for n ≥ 1
    3. Σ n/(n + 1) for n ≥ 1
    4. Σ 1/ln n for n ≥ 2
    Show answer

    Answer: B — Σ 1/n^1.5 for n ≥ 1

    Σ1/nᵖ converges iff p > 1. Here p = 1.5 > 1, so Σ1/n^1.5 converges. Σ1/√n has p = 1/2 and diverges. The terms n/(n + 1) tend to 1, not 0, so that series diverges by the necessary condition, and 1/ln n > 1/n makes Σ1/ln n diverge by comparison with the harmonic series.

  2. The series Σn≥1 n²/3ⁿ is examined with the ratio test. The limit of aₙ₊₁/aₙ and the conclusion are:

    1. 1, so the test fails and another test is needed
    2. 3, so the series diverges
    3. 1/3, so the series converges
    4. 1/9, so the series converges
    Show answer

    Answer: C — 1/3, so the series converges

    aₙ₊₁/aₙ = (n + 1)²/(3n²) = (1/3)(1 + 1/n)² → 1/3 < 1, so the series converges. The factor n² only contributes (1 + 1/n)² → 1; a limit of 1 would be the signature of a p-series such as Σ1/n², not of this exponential-type term.

  3. Cauchy's condensation test is applied to Σn≥2 1/(n (ln n)ᵖ). For which values of p does the series converge?

    1. p > 1 only
    2. p ≥ 1
    3. p > 0
    4. p < 1 only
    Show answer

    Answer: A — p > 1 only

    The condensed term is 2ᵏ·1/(2ᵏ (k ln 2)ᵖ) = 1/((ln 2)ᵖ kᵖ), a p-series in k. It converges iff p > 1. At p = 1 the series diverges (it is Σ1/(n ln n)), which excludes p ≥ 1 as the answer.

  4. The interval of convergence of Σn≥1 (x − 2)ⁿ/(n·3ⁿ) is:

    1. (−1, 5)
    2. [−1, 5]
    3. (−1, 5]
    4. [−1, 5)
    Show answer

    Answer: D — [−1, 5)

    R = 3 with centre 2 gives (−1, 5). At x = −1 the series is Σ(−1)ⁿ/n, convergent by the alternating test; at x = 5 it is Σ1/n, divergent. So the interval is [−1, 5), including the left end only. Forgetting the endpoint test gives (−1, 5).

  5. The Maclaurin series of ln(1 + x) converges for −1 < x ≤ 1. Using it, the sum of the series 1 − 1/2 + 1/3 − 1/4 + … is:

    1. 1
    2. ln 2
    3. π/4
    4. it diverges
    Show answer

    Answer: B — ln 2

    Putting x = 1, which is allowed because the right end is included, gives ln 2 = 1 − 1/2 + 1/3 − …. The series converges conditionally, by the alternating test. π/4 is the sum of the arctan series 1 − 1/3 + 1/5 − …, not of this one.

  6. A function f has f(0) = 1, f′(0) = 0, f″(0) = −2 and f‴(0) = 0. Its Maclaurin polynomial of degree 3 is P₃(x) = 1 − x². The Lagrange form of the error says that for |x| ≤ 0.5, if |f⁽⁴⁾| ≤ 48 everywhere, the error |f(x) − P₃(x)| is at most:

    1. 0.125
    2. 0.25
    3. 0.0625
    4. 0.03125
    Show answer

    Answer: A — 0.125

    |R₃| ≤ M|x|⁴/4! = 48 × (0.5)⁴/24 = 48 × 0.0625/24 = 0.125. The degree of the remainder is one more than that of the polynomial, so 4! and x⁴ are used, not 3! and x³ (which would give 48 × 0.125/6 = 1).

  7. Cauchy's mean value theorem is applied to f(x) = x² and g(x) = x³ on [1, 2]. The point c guaranteed by the theorem lies at:

    1. c = 3/2
    2. c = 14/9
    3. c = 7/4
    4. c = 9/7
    Show answer

    Answer: B — c = 14/9

    The quotient f′/g′ at c is 2c/(3c²) = 2/(3c) and must equal (f(2) − f(1))/(g(2) − g(1)) = 3/7, so c = 14/9. The midpoint 3/2 would be the Lagrange point for the single function x² (f′ = 2c = 3), a common slip, but here the quotient of derivatives is used.

  8. For a continuous f on [a, b], d/dx ∫ₐx³ f(t) dt equals:

    1. f(x³)
    2. 3x² f(x)
    3. 3x² f(x³)
    4. f(x³) − f(a)
    Show answer

    Answer: C — 3x² f(x³)

    By the fundamental theorem with the chain rule the result is f(u(x))·u′(x) with u = x³, i.e. 3x² f(x³). Writing f(x³) alone forgets u′; the expression f(x³) − f(a) confuses the integral with an antiderivative.

  9. A function is continuous on [0, 2] with f(x) = 1 for 0 ≤ x ≤ 1 and f(x) = 1 + (x − 1)² for 1 < x ≤ 2. By the mean value theorem of integral calculus there is a c in [0, 2] with f(c) equal to:

    1. 7/6
    2. 4/3
    3. 1
    4. 2/3
    Show answer

    Answer: A — 7/6

    The integral is ∫₀¹ 1 dx + ∫₁² (1 + (x − 1)²) dx = 1 + (1 + 1/3) = 7/3. Dividing by the length 2 gives the average 7/6, which lies between the minimum 1 and maximum 2 of f, so c exists. The value 4/3 is the integral over [1, 2] alone.

  10. Which of the following series converge?

    1. Σ n!/nⁿ
    2. Σ (n/(2n + 1))ⁿ
    3. Σ 1/(n ln n) for n ≥ 2
    4. Σ n/(n + 5) for n ≥ 1
    Show answer

    Answer: A — Σ n!/nⁿ; B — Σ (n/(2n + 1))ⁿ

    Σn!/nⁿ: ratio → 1/e < 1, converges. Σ(n/(2n + 1))ⁿ: root → 1/2 < 1, converges. Σ1/(n ln n) diverges by condensation (harmonic after condensing). Σn/(n + 5) has terms → 1, so it diverges by the necessary condition.

  11. Which statements about series and power series are correct?

    1. If Σ|aₙ| converges then Σaₙ converges
    2. If aₙ → 0 then Σaₙ converges
    3. Differentiating a power series term by term does not change its radius of convergence
    4. The ratio test always decides convergence when the limit is 1
    Show answer

    Answer: A — If Σ|aₙ| converges then Σaₙ converges; C — Differentiating a power series term by term does not change its radius of convergence

    Absolute convergence implies convergence, and term-by-term differentiation or integration keeps R (the endpoint behaviour may change). aₙ → 0 is only necessary, with Σ1/n as the counter-example. A ratio limit of 1 is inconclusive, as Σ1/n and Σ1/n² show.

  12. Which statements about the mean value theorems are correct?

    1. Lagrange's theorem is Cauchy's theorem with g(x) = x
    2. Rolle's theorem holds for f(x) = |x| on [−1, 1] because f(−1) = f(1)
    3. The mean value theorem of integral calculus needs f continuous on [a, b]
    4. If f′(x) = 0 for all x in (a, b) then f is constant on [a, b] (f continuous on [a, b])
    Show answer

    Answer: A — Lagrange's theorem is Cauchy's theorem with g(x) = x; C — The mean value theorem of integral calculus needs f continuous on [a, b]; D — If f′(x) = 0 for all x in (a, b) then f is constant on [a, b] (f continuous on [a, b])

    g(x) = x makes the Cauchy quotient (f(b) − f(a))/(b − a), which is Lagrange's statement. The integral mean value theorem needs continuity. Zero derivative gives constancy by the Lagrange theorem. Rolle's theorem fails for |x| because it is not differentiable at 0, even though f(−1) = f(1); there is no point with zero slope.

  13. The sum of the series Σn=1∞ n/3ⁿ is S. Find the value of 4S.

    Numerical answer — type the value.

    Show answer

    Answer: 3

    Σn xⁿ = x/(1 − x)² for |x| < 1. At x = 1/3: S = (1/3)/(2/3)² = (1/3)(9/4) = 3/4, so 4S = 3. Second check: summing n/3ⁿ for n up to 200 numerically gives 0.75000.

  14. The error of the Maclaurin polynomial of degree n for eˣ at x = 1 is bounded by e/(n + 1)!. Taking e < 3, find the smallest n for which the guaranteed error bound 3/(n + 1)! is below 10⁻⁴.

    Numerical answer — type the value.

    Show answer

    Answer: 7

    n = 6: 3/7! = 3/5040 = 5.95 × 10⁻⁴ > 10⁻⁴. n = 7: 3/8! = 3/40320 = 7.44 × 10⁻⁵ < 10⁻⁴. A Python loop over n = 1…11 gives the same first n = 7.

  15. Use the second-order Taylor polynomial of f(x) = √x about a = 4 to estimate √4.1. Give the answer to four decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 2.0248

    f(4) = 2, f′(4) = 1/4, f″(4) = −1/32. P₂(4.1) = 2 + 0.1/4 − (1/32)(0.01)/2 = 2 + 0.025 − 0.00015625 = 2.02484375 ≈ 2.0248. Direct √4.1 = 2.0248457, same to four places.

  16. Cauchy's mean value theorem is applied to f(x) = x² and g(x) = x³ on [1, 2]. If c is the point promised, find the value of 9c.

    Numerical answer — type the value.

    Show answer

    Answer: 14

    2c/(3c²) = 3/7 gives 2/(3c) = 3/7, so c = 14/9 and 9c = 14. Substituting back: 2/(3 × 14/9) = 18/42 = 3/7, as required.

  17. For the power series Σn≥0 (2n)! xⁿ/(n!)² the radius of convergence is R. Find 1/R.

    Numerical answer — type the value.

    Show answer

    Answer: 4

    aₙ₊₁/aₙ = (2n + 2)(2n + 1)/(n + 1)² = 2(2n + 1)/(n + 1) → 4. So R = 1/4 and 1/R = 4. Check at n = 1000: 2(2001)/1001 = 3.998.