به انگلیسی

How far a power series reaches

Mathematics · Calculus & Analysis · ages 19-20

Name ______________________ Date ____________

The nonzero coefficients satisfy |a_(n+1)/a_n| -> 1/4. What is the radius of convergence of sum a_n(x-2)^n, starting at n = 0?
Answer: ______________
For sum x^n/2^n, starting at n = 1, what is the radius of convergence?
Answer: ______________
A power series has positive radius R. The series obtained by differentiating it term by term has the same radius R.
Circle one: True False
For sum n! x^n, starting at n = 1, what is the radius of convergence?
- 1
- Infinite
- 0
Integrate sum x^n, starting at n = 0, term by term from 0 to x. What is the radius of convergence of the resulting series?
Answer: ______________
F(x) = sum x^n/n, starting at n = 1, has radius 1. Differentiate term by term inside that radius. What is the full interval of convergence of the differentiated series?
- (-1, 1)
- [-1, 1)
- (-1, 1]
- [-1, 1]
For sum (x-2)^n/n!, starting at n = 1, which statement describes its convergence?
- Only x = 2 gives convergence.
- Every real x gives convergence.
- Only |x-2| < 1 gives convergence.
- Only |x-2| <= 1 gives convergence.
A power series centered at 3 has radius 1. It converges at x = 2 and diverges at x = 4. What is its interval of convergence?
- (2, 4)
- [2, 4)
- (2, 4]
Sam says every power series with radius 1 converges at both endpoints. Is Sam's claim true?
Circle one: True False
For sum (2x)^n/n, starting at n = 1, what is the radius of convergence? Give a decimal.
Answer: ______________

Answer key

For grown-ups. Fold this page away before handing over the rest.

4 · The term-ratio limit is |x-2|/4. It is below one when |x-2| < 4, so the radius is 4.
2 · Away from zero the absolute ratio is |x|/2. The condition |x|/2 < 1 gives |x| < 2, so R = 2. The series also converges at zero.
True · Term-by-term differentiation preserves the radius. It need not preserve convergence at the endpoints.
0 · For nonzero x, the ratio (n+1)|x| tends to infinity and proves divergence. At x = 0 every term vanishes. Only the center converges, so R = 0.
1 · The original geometric series converges exactly for |x| < 1, so its radius is 1. Integration gives terms x^(n+1)/(n+1) and preserves that radius.
(-1, 1) · The differentiated terms are x^(n-1), with radius 1. At x = 1 all terms are 1; at x = -1 their magnitudes are 1. Neither endpoint converges, so the interval is (-1, 1).
Every real x gives convergence. · For every fixed x other than 2, the absolute ratio |x-2|/(n+1) tends to zero. At x = 2 all terms vanish. Thus every real x gives convergence.
[2, 4) · The interior is (3-1, 3+1) = (2, 4). Include 2 because it converges and exclude 4 because it diverges. The interval is [2, 4).
False · The geometric power series sum x^n, starting at n = 0, has radius 1. At x = 1 and x = -1 its terms do not tend to zero, so both endpoints diverge.
0.5 · The absolute ratio is 2|x|n/(n+1), whose limit is 2|x|. Solving 2|x| < 1 gives |x| < 0.5, so R = 0.5.