Master the culmination of AP Calculus BC with convergence, power series, and Taylor polynomials
← Back to UnitsUnit 10 introduces the fascinating world of infinite sequences and series, where we explore convergence, power series, and Taylor polynomials. These concepts are essential for understanding advanced mathematics and have applications in physics, engineering, and computer science.
A sequence is an ordered list of numbers: {a₁, a₂, a₃, ...} or {aₙ}
Determine if the sequence aₙ = (n² + 1)/(2n² - 3) converges.
Solution:
1) limn→∞ (n² + 1)/(2n² - 3)
2) Divide by highest power: limn→∞ (1 + 1/n²)/(2 - 3/n²)
3) = (1 + 0)/(2 - 0) = 1/2
4) The sequence converges to 1/2.
An infinite series is the sum of an infinite sequence:
The series converges if the sequence of partial sums Sₙ = a₁ + a₂ + ... + aₙ converges.
Determine if ∑n=0∞ (1/3)ⁿ converges and find its sum.
Solution:
1) This is a geometric series with a = 1 and r = 1/3
2) Since |r| = 1/3 < 1, the series converges
3) Sum = a/(1-r) = 1/(1-1/3) = 1/(2/3) = 3/2
Test ∑n=1∞ n!/nⁿ for convergence.
Solution:
1) aₙ = n!/nⁿ
2) aₙ₊₁ = (n+1)!/(n+1)ⁿ⁺¹
3) |aₙ₊₁/aₙ| = [(n+1)!/(n+1)ⁿ⁺¹] / [n!/nⁿ]
4) = (n+1) · nⁿ / (n+1)ⁿ⁺¹ = nⁿ / (n+1)ⁿ
5) = (n/(n+1))ⁿ = (1/(1+1/n))ⁿ → 1/e < 1
6) Since limit < 1, series converges by Ratio Test
A power series centered at x = a is:
Find the radius of convergence for ∑n=0∞ xⁿ/n!
Solution:
1) cₙ = 1/n!
2) |cₙ₊₁/cₙ| = (1/(n+1)!) / (1/n!) = n!/(n+1)! = 1/(n+1)
3) R = limn→∞ |cₙ/cₙ₊₁| = limn→∞ (n+1) = ∞
4) Radius of convergence is ∞ (converges for all x)
The Taylor series for f(x) centered at x = a is:
When a = 0, this becomes the Maclaurin series.
Find the Maclaurin series for f(x) = eˣ.
Solution:
1) f(x) = eˣ, so f⁽ⁿ⁾(x) = eˣ for all n
2) f⁽ⁿ⁾(0) = e⁰ = 1 for all n
3) eˣ = ∑n=0∞ (1/n!) xⁿ
4) = 1 + x + x²/2! + x³/3! + x⁴/4! + ...
The nth-degree Taylor polynomial is the first n+1 terms of the Taylor series:
Find the 3rd-degree Taylor polynomial for f(x) = ln(x) centered at x = 1.
Solution:
1) f(x) = ln(x), f(1) = 0
2) f'(x) = 1/x, f'(1) = 1
3) f''(x) = -1/x², f''(1) = -1
4) f'''(x) = 2/x³, f'''(1) = 2
5) P₃(x) = 0 + 1(x-1) + (-1/2)(x-1)² + (2/6)(x-1)³
6) = (x-1) - (x-1)²/2 + (x-1)³/3
Determine if aₙ = (3n² + 2n)/(n² + 1) converges.
Solution:
1) limn→∞ (3n² + 2n)/(n² + 1)
2) Divide by n²: limn→∞ (3 + 2/n)/(1 + 1/n²)
3) = (3 + 0)/(1 + 0) = 3
4) The sequence converges to 3.
Test ∑n=1∞ 1/(n² + 1) for convergence.
Solution:
1) Compare to p-series ∑1/n² (converges since p = 2 > 1)
2) 1/(n² + 1) < 1/n² for all n ≥ 1
3) By Direct Comparison Test, since ∑1/n² converges, ∑1/(n² + 1) also converges
Find the radius of convergence for ∑n=0∞ (x-2)ⁿ/n.
Solution:
1) cₙ = 1/n
2) |cₙ₊₁/cₙ| = (1/(n+1)) / (1/n) = n/(n+1)
3) R = limn→∞ |cₙ/cₙ₊₁| = limn→∞ (n+1)/n = 1
4) Radius of convergence is 1
Now that you've mastered infinite sequences and series, test your understanding with our comprehensive practice test. The test includes: