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Differences, reduction formulae and limits

Sum series by the method of differences, derive reduction formulae, evaluate improper integrals and compare growth rates.

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  • GCE A-Level H3 Mathematics 2027
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Enlarged diagram

Three H3 tools in this lesson share one idea: replace an infinite or repeated process by something finite you can control. The method of differences collapses a long sum to a few surviving terms. A reduction formula steps an integral down one index at a time until it can be evaluated. An improper integral is defined as a limit of ordinary integrals, and comparing growth rates decides whether that limit exists. Each section ends with the check that examiners reward: test the formula at a small case, or test the limit numerically.

The method of differences

If each term of a series can be written as a difference

uᵣ = f(r)-f(r + k),

then writing the sum out row by row makes most terms cancel in pairs, leaving k terms at the start and k at the end. The work is in finding f; partial fractions or a factorial identity usually provides it.

Worked example 1

A sum that collapses

Find ∑ᵣ₌₁ⁿ1/(r(r + 2)) and its limit as n → ∞.

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  1. Split the term

    Method

    By partial fractions, 1/(r(r + 2)) = (1/2)(1/r-1/(r + 2)).

    Reason

    This is a difference f(r)-f(r + 2) with f(r) = 1/(2r), so the gap is 2.

    Working

    Answer

    Write the rows out, keeping the factor 1/2 outside:

    r = 1: 1/1-1/3; r = 2: 1/2-1/4; r = 3: 1/3-1/5; ⋮; r = n-1: 1/(n-1)-1/(n + 1); r = n: 1/n-1/(n + 2)

    Each 1/r with r ≥ 3 cancels with the -1/r two rows earlier.

  2. Keep what survives

    Method

    Two positive terms survive at the start and two negative terms at the end.

    Reason

    A gap of 2 leaves two terms at each end.

    Working

    Answer

    ∑ᵣ₌₁ⁿ1/(r(r + 2)) = (1/2)(1 + 1/2-1/(n + 1)-1/(n + 2)); = 3/4-(2n + 3)/(2(n + 1)(n + 2)).

    As n → ∞ the last term tends to 0, so the sum to infinity is 3/4.

Guided practice 1

A difference hidden in a factorial

About 6 min

Show that r · r! = (r + 1)!-r!, and hence find ∑ᵣ₌₁ⁿr · r!. Use your formula to find the sum when n = 5.

Try this before viewing the solution

Hint 1

Factor the right-hand side

(r + 1)! = (r + 1) r!, so (r + 1)!-r! = r! ((r + 1)-1).

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  1. Prove the identity

    Method

    Factor r! out of the right-hand side.

    Reason

    The identity writes each term as f(r + 1)-f(r) with f(r) = r!, a gap of 1.

    Working

    Answer

    (r + 1)!-r! = r! (r + 1-1) = r · r!.

  2. Cancel and evaluate

    Working

    Answer

    ∑ᵣ₌₁ⁿr · r! = (2!-1!) + (3!-2!) + … + ((n + 1)!-n!); = (n + 1)!-1.

    For n = 5 the sum is 6!-1 = 719.

Reduction formulae step down an index

A reduction formula links an integral that depends on n to the same integral with a smaller index, usually through integration by parts. Applying it repeatedly reaches an integral you can evaluate directly.

Worked example 2

Powers of sine

Let Sₙ = ∫₀^(π/2) sinⁿ x dx for n ≥ 0. Show that Sₙ = ((n-1)/n)Sₙ₋₂ for n ≥ 2, and hence find S₄ and S₅.

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  1. Integrate by parts

    Method

    Write sinⁿ x = sinⁿ⁻¹ x · sin x, and take u = sinⁿ⁻¹ x, v' = sin x.

    Reason

    Differentiating sinⁿ⁻¹ x lowers the power of sine, and the boundary term vanishes at both limits.

    Working

    Answer

    Sₙ = [- sinⁿ⁻¹ x cos x]₀^(π/2) + (n-1)∫₀^(π/2) sinⁿ⁻² x cos² x dx; = 0 + (n-1)∫₀^(π/2) sinⁿ⁻² x(1- sin² x)dx; = (n-1)Sₙ₋₂-(n-1)Sₙ.

  2. Solve for the original integral

    Method

    Collect the two Sₙ terms.

    Reason

    The original integral reappears on the right; treat it as an unknown rather than integrating again.

    Working

    Answer

    nSₙ = (n-1)Sₙ₋₂ ⇒ Sₙ = ((n-1)/n)Sₙ₋₂.

  3. Step down to a known integral

    Working

    Answer

    S₀ = π/2 and S₁ = 1, so

    S₄ = 3/4 · 1/2 · π/2 = 3π/16,; S₅ = 4/5 · 2/3 · 1 = 8/15.

Check your understanding 1

Powers times an exponential

Let Iₙ = ∫₀¹xⁿe^x dx for n ≥ 0. Show that Iₙ = e-nIₙ₋₁ for n ≥ 1, and hence find I₃ exactly.

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By parts with u = xⁿ, v' = e^x:

Iₙ = [xⁿe^x]₀¹-n∫₀¹xⁿ⁻¹e^x dx = e-nIₙ₋₁ (n ≥ 1).

I₀ = e-1, so I₁ = 1, I₂ = e-2 and I₃ = e-3(e-2) = 6-2e ≈ 0.563. This is positive, as an integral of a positive function must be.

Improper integrals are limits

An integral with an infinite limit, or with an integrand that is unbounded at an end point, is improper. It is defined as a limit, and it converges only if that limit exists:

∫₁^∞ f(x) dx = lim _(b → ∞)∫₁^bf(x) dx.

The graphs of 1/x and 1/x² look alike, but their areas behave very differently:

Upper limit b∫₁^bdx/x² = 1-1/b∫₁^bdx/x = ln b
100.92.30
1000.994.61
10000.9996.91

The first settles towards 1; the second keeps growing without bound, slowly but forever.

Growth rates decide limits

Limits obey the familiar rules when the separate limits exist: the limit of a sum is the sum of the limits, and similarly for differences, products and quotients with a non-zero limit below. When a quotient has both parts growing, compare growth rates. As x → ∞, every positive power of x eventually beats every power of ln x, and e^x beats every power of x:

x^k/(e^x) → 0, (ln x)/(x^k) → 0 (k > 0).

“Eventually” matters: x³/(e^x) is about 0.84 at x = 5 but only 0.045 at x = 10 and 1.6 × 10⁻⁵ at x = 20.

Check your understanding 2

Converge or diverge?

Decide whether each integral converges, and find its value if it does.

  1. ∫₁^∞1/x² dx
  2. ∫₁^∞1/x dx
  3. ∫₀¹1/(square root of x) dx
  4. ∫₀^∞ xe^(-x) dx
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  1. ∫₁^b x⁻² dx = 1-1/b → 1: converges to 1.
  2. ∫₁^b x⁻¹ dx = ln b → ∞: diverges.
  3. ∫ₐ¹x^(-1/2) dx = 2-2 square root of a → 2 as a → 0⁺: converges to 2, although the integrand is unbounded.
  4. By parts, ∫₀^bxe^(-x) dx = 1-(b + 1)e^(-b) → 1, because (b + 1)/(e^b) → 0: converges to 1.

Common mistakes

  • Keeping one surviving term at each end when the gap k is greater than 1.
  • Concluding that a series or improper integral converges just because its terms tend to 0: ∫₁^∞dx/x diverges although 1/x → 0.
  • In a reduction formula, forgetting the boundary term from integration by parts, or not checking that it is zero.
  • Stating the value of an improper integral without writing it as a limit, or treating ∞ as a number to substitute.
  • Splitting a limit into separate limits when one of them does not exist, as in lim(x- square root of (x² + x)).

Before you move on

Check your understanding 3

Two convergent integrals

Find the value of each improper integral.

  1. ∫₀^∞ x²e^(-x) dx
  2. ∫₁^∞(ln x)/x² dx
Show answerHide answer
  1. With Jₙ = ∫₀^∞ xⁿe^(-x) dx, integration by parts gives Jₙ = lim _(b → ∞)[-xⁿe^(-x)]₀^b + nJₙ₋₁ = nJₙ₋₁, since bⁿ/(e^b) → 0. With J₀ = 1, J₂ = 2 × 1 × 1 = 2.
  2. By parts,
∫₁^b(ln x)/x² dx = [-(ln x)/x-1/x]₁^b = 1-(ln b)/b-1/b → 1,

because (ln b)/b → 0.

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