Radius and Interval of Convergence of Power Series: Practice Questions (Calculus BC 10.13)

Answer: (C). Ratio of consecutive terms: |uₙ₊₁/uₙ| = ((n + 1)/n) · |x − 4|/5 → |x − 4|/5. Setting |x − 4|/5 < 1 gives |x − 4| < 5, so R = 5.

• (A) takes the limit of the coefficient ratio, (n + 1)/(5n) → 1/5, and calls that the radius. That limit is 1/R, not R.
• (B) is the centre of the series, not its radius.
• (D) is the length of the open interval (−1, 9). The radius is half of that.

Answer: (C). The ratio test gives |x + 1|/2 · √(n/(n + 1)) → |x + 1|/2, so |x + 1| < 2: centre −1, radius 2, open interval (−3, 1).

• At x = 1: x + 1 = 2, so the series is Σ (−1)ⁿ/√n. The terms alternate, 1/√n decreases and tends to 0, so it converges by the alternating series test.
• At x = −3: x + 1 = −2, so the term is (−1)ⁿ(−2)ⁿ/(√n · 2ⁿ) = (+1)ⁿ/√n = 1/√n. This is a p-series with p = ½ ≤ 1, so it diverges.

So the interval is (−3, 1].

• (A) stops at the open interval and never tests the endpoints.
• (B) forgets that the (−1)ⁿ in the series combines with (−2)ⁿ at x = −3, so it swaps which endpoint alternates.
• (D) treats Σ 1/√n as convergent. A p-series converges only when p > 1.

Answer: (B). For n! xⁿ, |uₙ₊₁/uₙ| = (n + 1)|x|. For any x ≠ 0 this tends to ∞, so the series diverges. At x = 0 every term after the first is 0, so it converges there. R = 0.

• (A) has ratio |x|/(n + 1) → 0 for every x, so it converges for all x (R = ∞). The factorial is in the denominator.
• (C) has ratio |x| · (n + 1)/(n + 2) → |x|, so R = 1.
• (D) is geometric with ratio 2x, so it converges for |x| < 1/2 (R = 1/2).

Answer: (C). A power series with positive radius is the Taylor series of its sum, so aₙ = f⁽ⁿ⁾(3)/n!, which means f⁽ⁿ⁾(3) = n! · aₙ. The coefficient of (x − 3)³ is a₃ = 4/2³ = 1/2. So f‴(3) = 3! · 1/2 = 3.

• (A) divides the coefficient by 3! instead of multiplying: (1/2)/6.
• (B) is the coefficient a₃ itself, not the derivative.
• (D) takes only the factor n + 1 = 4 and forgets the 2³ in the denominator and the 3!.

1. Ratio. |uₙ₊₁/uₙ| = |x^(2n+2)| / (4ⁿ⁺¹(n + 2)) · (4ⁿ(n + 1)) / |x^(2n)| = (x²/4) · (n + 1)/(n + 2).
2. Limit. (n + 1)/(n + 2) → 1, so the limit is x²/4.
3. Solve. x²/4 < 1 gives x² < 4, so |x| < 2. R = 2, centre 0.
4. Endpoints. At x = 2 or x = −2, x^(2n) = 4ⁿ (the power is even, so the sign does not matter). Each series becomes Σ 1/(n + 1) = 1 + ½ + ⅓ + …, the harmonic series, which diverges.

Interval of convergence: (−2, 2).

Suggested mark points (3): 1 for a correct ratio with limit x²/4; 1 for R = 2; 1 for testing both endpoints, naming the harmonic series, and giving (−2, 2).

Common error: treating x^(2n) as xⁿ and getting |x| < 4, or writing R = 4 from x² < 4.

(a) |uₙ₊₁/uₙ| = (|x − 3|/2) · n²/(n + 1)² → |x − 3|/2. Converges when |x − 3| < 2, so R = 2.

(b) Open interval (1, 5).

• x = 5: Σ 2ⁿ/(n² 2ⁿ) = Σ 1/n², a p-series with p = 2 > 1. Converges.
• x = 1: Σ (−2)ⁿ/(n² 2ⁿ) = Σ (−1)ⁿ/n². Its absolute values form Σ 1/n², which converges, so it converges absolutely.

Interval of convergence: [1, 5].

(c) f′(x) = Σ from n = 1 to ∞ of n(x − 3)ⁿ⁻¹/(n² 2ⁿ) = Σ from n = 1 to ∞ of (x − 3)ⁿ⁻¹/(n · 2ⁿ). Its radius is 2, because a power series differentiated term by term has the same radius of convergence as the original.

(d)

• x = 5: Σ 2ⁿ⁻¹/(n 2ⁿ) = Σ 1/(2n) = ½ Σ 1/n, a multiple of the harmonic series. Diverges.
• x = 1: Σ (−2)ⁿ⁻¹/(n 2ⁿ) = Σ (−1)ⁿ⁻¹/(2n). Alternating, 1/(2n) decreases to 0, so it converges (alternating series test).

Interval of convergence of f′: [1, 5).

Total: 6 points. Acceptable alternative for (c): a fresh ratio test on the f′ series, which also gives R = 2. In (d), “the endpoints are the same as for f” earns no credit: the right endpoint is lost.

(a) x = 8 is 3 units from the centre 5. A power series diverges whenever |x − 5| > R, so converging at distance 3 means 3 ≤ R. x = 1 is 4 units from the centre, and the series converges whenever |x − 5| < R, so diverging at distance 4 means R ≤ 4. 3 ≤ R ≤ 4.

(b)

• x = 7.5: distance 2.5 < 3 ≤ R, so it is strictly inside the radius. Must converge.
• x = 10: distance 5 > 4 ≥ R. Must diverge.
• x = 2: distance 3. If R = 3, x = 2 is an endpoint and could go either way; if R > 3 it is inside. Cannot tell.

(c) Ratio: |uₙ₊₁/uₙ| = (|x − 5|/3) · n/(n + 1) → |x − 5|/3, so R = 3 (which fits 3 ≤ R ≤ 4).

• x = 8: x − 5 = 3, so the term is (−1)ⁿ3ⁿ/(n 3ⁿ) = (−1)ⁿ/n. Converges (alternating series test). ✓
• x = 1: x − 5 = −4, so |term| = (4/3)ⁿ/n, which tends to ∞. Diverges (nth term test). ✓
• x = 2: x − 5 = −3, so the term is (−1)ⁿ(−3)ⁿ/(n 3ⁿ) = 1/n. Harmonic series: diverges.

Interval of convergence: (2, 8].

Total: 6 points. Note for (b): a different series, such as aₙ = 1/(n² 3ⁿ), also converges at 8 and diverges at 1 but converges at x = 2. That is why “cannot tell” is the right answer there.