The geometric sequence aₙ = 3·0.5ⁿ⁻¹ decreases. Find the first index n for which aₙ < 0.5.
Sequences & series worksheet with answers
1.Easy
a₁ = 3r = 0.5ε = 0.5
n =
2.Easy
Find the sum S of the infinite series Σk=1∞ 24·(−2/5)ᵏ.
a = 24r = −2/5k₀ = 1
S =
3.Medium
Apply the ratio test to Σk=1∞ k!/kᵏ. Find L = lim |ak+1/ak|.
ak = k!/kᵏ
L =
4.Medium
Exactly one of the four series diverges. Click it.
A: Σ 1/(k(k + 3))B: Σ 1/(2k + 6)C: Σ k²/5ᵏD: Σ 1/k⁴
- 1A: Σ 1/(k(k + 3))
- 2B: Σ 1/(2k + 6)
- 3C: Σ k²/5ᵏ
- 4D: Σ 1/k⁴
5.Hard
f(x) = sin x is approximated by its Maclaurin polynomial of degree 2 at x = 0.2. Use the Lagrange error bound M·|x|ⁿ⁺¹/(n + 1)! with M = 1 to find the largest possible error |R₂|.
f(x) = sin xn = 2x = 0.2M = 1
|R₂| =
6.Hard
f(x) = eˣ is approximated at x = 0.5 by its Maclaurin polynomial of degree n. With the Lagrange bound M·|x|ⁿ⁺¹/(n + 1)! and M = 2, find the smallest degree n that guarantees an error below 0.0001.
f(x) = eˣx = 0.5M = 2ε = 0.0001
n =
Answer key
- 1. n = 4
- 2. S = −6.857
- 3. L = 0.3679
- 4. B: Σ 1/(2k + 6)
- 5. |R₂| = 0.001333
- 6. n = 5
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