Questions: Integral Test

Question 1

You apply the integral test and find that ∫₁^∞ f(x)dx = 7. What can you conclude about the series Σ f(n)?

• A. The series converges and its sum equals 7
• B. The series converges, but its sum is not necessarily 7
• C. The series diverges because the integral value must match the series sum exactly
• D. Nothing — the integral test only applies when the integral diverges

Answer: (B)

The integral test tells you only whether the series converges or diverges — it does not give the series' sum. If the improper integral converges (finite value), the series also converges; if the integral diverges, so does the series. But the integral's value (7) is not the series' sum — they differ by a finite amount bounded by f(1). For example, ∫₁^∞ 1/x² dx = 1 but Σ 1/n² = π²/6 ≈ 1.645. The most common mistake is treating the integral test as computing the series value.

Question 2

Which series does the integral test most directly establish as convergent?

• A. Σ (-1)ⁿ/n — alternating terms, not always positive
• B. Σ 1/n² — positive, continuous, decreasing; ∫₁^∞ 1/x² dx converges
• C. Σ n·sin(n) — oscillating, not eventually monotone decreasing
• D. Σ 1/n — the harmonic series, whose integral also diverges

Answer: (B)

For Σ 1/n², set f(x) = 1/x². This function is positive, continuous, and decreasing on [1, ∞). The integral ∫₁^∞ 1/x² dx = [-1/x]₁^∞ = 1, which converges, so the series converges. Σ (-1)ⁿ/n fails the positivity condition (the integral test requires non-negative terms). Σ n·sin(n) is not monotone decreasing. Σ 1/n satisfies the conditions but has a divergent integral (∫₁^∞ 1/x dx diverges), making it an example of divergence, not convergence.

Question 3

If ∫₁^∞ f(x)dx converges to a finite value L, then Σ f(n) also converges to L.

• T. True
• F. False

Answer: False

The integral test establishes only that the series and integral share the same convergence behavior — both converge or both diverge. Their numerical values are almost never equal. The rectangle picture makes this clear: the series terms aₙ = f(n) are rectangle areas that differ from the integral by a finite amount (bounded above by f(1) in typical estimates). For example, ∫₁^∞ 1/x² dx = 1 while Σ 1/n² = π²/6 ≈ 1.645. Confusing the integral's value with the series' sum is the single most common misapplication of this test.

Question 4

The integral test can be applied to determine convergence of Σ sin(n)/n² because f(x) = sin(x)/x² is eventually decreasing.

• T. True
• F. False

Answer: False

The integral test requires f to be positive (as well as continuous and decreasing). The function sin(x)/x² takes negative values whenever sin(x) < 0, so the positivity condition fails. Without positivity, the rectangle-to-curve comparison breaks down: rectangles below the x-axis would subtract from the sum rather than add, and the geometric argument for convergence equivalence no longer holds. For series with sign changes, the alternating series test or absolute convergence approach is more appropriate.

Question 5

Explain geometrically why a series Σ f(n) and the improper integral ∫₁^∞ f(x)dx share the same convergence behavior when f is positive, continuous, and decreasing.

Model answer: Draw the graph of f and place rectangles of width 1 over each integer n, with height f(n). Because f is decreasing, each rectangle on [n, n+1] with height f(n) lies above the curve, making the series an overestimate of the integral: ∫₁^∞ f(x)dx ≤ Σ_{n=1}^∞ f(n). Using heights f(n+1) instead gives rectangles below the curve: Σ_{n=2}^∞ f(n) ≤ ∫₁^∞ f(x)dx. These two inequalities sandwich the integral between two shifted versions of the series, showing they differ by at most a finite amount (f(1)). Therefore, if one diverges to infinity, the other must too; if one is bounded, the other is as well.

The decreasing condition is essential: it ensures the rectangle at n lies consistently above (or below) the curve on [n, n+1], giving a one-sided bound. Without monotone decrease, the rectangle could cross the curve, and you could no longer bound the series from one side by the integral. This is why 'eventually decreasing' is sufficient — behavior at finitely many early terms contributes only a finite amount and cannot affect convergence.