En anglais

When signs keep switching

Mathematics · Calculus & Analysis · ages 19-20

Name ______________________ Date ____________

A series has magnitudes 1/n, n ≥ 1, and signs repeating +,+,-,-. What does the alternating series test tell you?
- It proves convergence because the magnitudes decrease to 0
- It does not apply because consecutive signs can be the same
- It proves divergence because consecutive signs can be the same
- It does not apply because the magnitudes stay away from 0
Why does Σ (-1)^(n+1)/3^n, summed from n = 1 to infinity, converge?
- Its signs alternate and its magnitudes decrease to 0
- Its terms stay positive and their magnitudes decrease to 0
- Its signs alternate and its absolute-term ratio tends to 0
The alternating series test allows equal consecutive magnitudes, provided the magnitudes are nonincreasing and tend to 0.
Circle one: True False
For terms a_n = (-1)^n/n⁵, n ≥ 1, what is p in the series of absolute values?
Answer: ______________
For Σ (-1)^(n+1)/n², n ≥ 1, what error bound does the alternating remainder theorem give after four terms? Enter a decimal.
Answer: ______________
For 1 - 1/2 + 1/3 - 1/4 + ..., what is the smallest number of terms that the alternating remainder bound certifies for error at most 0.025?
- 39 terms
- 38 terms
- 40 terms
- 41 terms
For n ≥ 1, each c_n is either 1 or -1, with no required sign pattern. What follows for Σ c_n/(n²+7)?
- It converges only if the signs alternate at every step
- It diverges whenever the signs stay positive at every step
- It converges absolutely for any allowed sign pattern
- It converges conditionally for any allowed sign pattern
All sums run from n = 1 to infinity. Which series converges conditionally?
- Σ (-1)^n/n⁶
- Σ (-1)^n/(n+4)
- Σ (-1)^n/4^n
What error bound does the alternating remainder theorem give after ten terms of 1 - 1/2 + 1/3 - 1/4 + ...?
- 1/11
- 1/10
- 1/100
A series has terms +2/k, -2/k in that order for each k ≥ 1. A student says equal consecutive magnitudes prevent convergence. Which partial-sum argument settles the claim?
- Even sums tend to 0, but odd sums tend to 2, so the series diverges
- Even sums are 0; odd sums are 2/k, so both tend to 0
- Every pair cancels, so odd partial sums also equal 0 for every positive integer k

Answer key

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

When signs keep switching W1-mt_sk6F4n1USH-s1

It does not apply because consecutive signs can be the same · The magnitudes decrease to 0, but the signs do not alternate at every term. This alternating series test does not apply.
Its signs alternate and its magnitudes decrease to 0 · The signs alternate. The magnitudes 1/3^n decrease to 0, so the alternating series test proves convergence.
True · The condition b_(n+1) ≤ b_n allows equality. Strict decrease at every step is not required.
5 · Taking absolute values gives 1/n⁵. Its exponent in the p-series form 1/n^p is 5.
0.04 · The first omitted magnitude is 1/5² = 1/25 = 0.04. The error is at most 0.04.
39 terms · Require 1/(N+1) ≤ 0.025 = 1/40. This gives N ≥ 39. At N = 38, the bound 1/39 exceeds 0.025.
It converges absolutely for any allowed sign pattern · The absolute terms are 1/(n²+7) ≤ 1/n². The convergent p-series proves absolute convergence by comparison.
Σ (-1)^n/(n+4) · Σ (-1)^n/(n+4) converges by the alternating series test. Its absolute series starts 1/5 + 1/6 + ..., a divergent harmonic tail. Its convergence is conditional.
1/11 · The first omitted term has index 11 and magnitude 1/11. The theorem bounds the error by that magnitude.
Even sums are 0; odd sums are 2/k, so both tend to 0 · Each completed pair cancels, so S_(2k) = 0. One unpaired term remains in S_(2k-1) = 2/k, which tends to 0. Thus all partial sums tend to 0.