Series 2

0.0(0)
Studied by 0 people
call kaiCall Kai
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
GameKnowt Play
Card Sorting

1/40

encourage image

There's no tags or description

Looks like no tags are added yet.

Last updated 6:33 AM on 10/9/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai
Chat

No analytics yet

Send a link to your students to track their progress

41 Terms

1
New cards
What is the necessary condition for the convergence of a series Σaₙ?
If Σaₙ converges, then aₙ → 0 as n → ∞.
2
New cards
What is the contrapositive of the necessary condition for convergence?
If aₙ does not tend to 0, then the series Σaₙ does not converge.
3
New cards
Does aₙ → 0 guarantee that Σaₙ converges?
No. It is necessary but not sufficient: a series may still diverge even though its terms tend to 0.
4
New cards
Why must aₙ → 0 when Σaₙ converges?
If sₙ is the sequence of partial sums, then aₙ = sₙ₊₁ − sₙ. If sₙ → S, then aₙ → S − S = 0.
5
New cards
What does it mean for a series to have eventually non-negative terms?
There exists an index N such that aₙ ≥ 0 for every n ≥ N.
6
New cards
What is the regularity theorem for series with eventually non-negative terms?
Every series with eventually non-negative terms is either convergent or positively divergent to +∞.
7
New cards
When does a series with eventually non-negative terms converge?
It converges if and only if its sequence of partial sums is bounded above.
8
New cards
Why are the partial sums of a series with non-negative terms increasing?
Because sₙ₊₁ = sₙ + aₙ₊₁ and aₙ₊₁ ≥ 0, so sₙ₊₁ ≥ sₙ.
9
New cards
What happens if the partial sums of a non-negative-term series are unbounded above?
The partial sums tend to +∞, so the series is positively divergent.
10
New cards
Why does changing or removing finitely many terms not affect a series’ character?
It changes the partial sums by only a fixed finite amount, so it cannot change convergence into divergence or vice versa.
11
New cards
What is the generalized harmonic series?
Σ from n=1 to ∞ of 1/n^α, where α ∈ ℝ.
12
New cards
When does the generalized harmonic series Σ from n=1 to ∞ of 1/n^α converge?
It converges when α > 1.
13
New cards
When does the generalized harmonic series Σ from n=1 to ∞ of 1/n^α diverge?
It diverges when α ≤ 1.
14
New cards
What is the generalized logarithmic harmonic series?
Σ from n=2 to ∞ of 1/[n^α(ln n)^β], where α, β ∈ ℝ.
15
New cards
When does Σ from n=2 to ∞ of 1/[n^α(ln n)^β] converge?
It converges when α > 1 and β > 0, and also when α = 1 and β > 1.
16
New cards
When does Σ from n=2 to ∞ of 1/[n^α(ln n)^β] diverge?
It diverges when α < 1 for every β ∈ ℝ, and also when α = 1 and β ≤ 1.
17
New cards
What does the generalized logarithmic harmonic series become when β = 0?
It becomes the generalized harmonic series Σ from n=2 to ∞ of 1/n^α.
18
New cards
What are the assumptions of the direct comparison criterion?
Σaₙ and Σbₙ have positive terms, and aₙ ≤ bₙ eventually.
19
New cards
What does the direct comparison criterion say about divergence?
If 0 < aₙ ≤ bₙ eventually and Σaₙ diverges to +∞, then Σbₙ also diverges to +∞.
20
New cards
What does the direct comparison criterion say about convergence?
If 0 < aₙ ≤ bₙ eventually and Σbₙ converges, then Σaₙ also converges.
21
New cards
Which series should be the larger one when using direct comparison to prove convergence?
The known convergent benchmark series should be the larger series: 0 ≤ aₙ ≤ bₙ and Σbₙ converges.
22
New cards
Which series should be the smaller one when using direct comparison to prove divergence?
The known divergent benchmark series should be the smaller series: 0 ≤ aₙ ≤ bₙ and Σaₙ diverges.
23
New cards
What does aₙ ~ bₙ mean?
It means lim as n → ∞ of aₙ/bₙ = 1; the sequences have the same dominant asymptotic behavior.
24
New cards
What is the asymptotic comparison criterion?
If aₙ > 0, bₙ > 0, and aₙ ~ bₙ, then Σaₙ and Σbₙ have the same character: either both converge or both diverge.
25
New cards
What does bₙ = o(aₙ) mean?
It means lim as n → ∞ of bₙ/aₙ = 0; bₙ is negligible compared with aₙ.
26
New cards
What proposition connects little-o notation with convergence?
If bₙ = o(aₙ) and Σaₙ converges, then Σbₙ converges.
27
New cards
What is the value of the exponential series Σ from n=0 to ∞ of 1/n!?
Σ from n=0 to ∞ of 1/n! = e.
28
New cards
What is the power-series representation of eˣ?
For every x ∈ ℝ, eˣ = Σ from n=0 to ∞ of xⁿ/n!.
29
New cards
What limit also represents eˣ?
For every x ∈ ℝ, eˣ = lim as n → ∞ of (1 + x/n)ⁿ.
30
New cards
How can a series with eventually non-positive terms be studied?
Rewrite Σaₙ as −Σ(−aₙ). Since −aₙ is eventually non-negative, use the criteria for non-negative series.
31
New cards
What does it mean for a series Σaₙ to converge absolutely?
Σaₙ converges absolutely if the series of absolute values Σ|aₙ| converges.
32
New cards
What is the absolute convergence criterion?
If Σ|aₙ| converges, then Σaₙ converges.
33
New cards
Does ordinary convergence imply absolute convergence?
No. A series may converge without converging absolutely.
34
New cards
What does it mean for a series to be conditionally convergent?
Σaₙ is conditionally convergent if Σaₙ converges but Σ|aₙ| diverges.
35
New cards
What is the alternating harmonic series?
Σ from n=1 to ∞ of (−1)ⁿ⁺¹/n = 1 − 1/2 + 1/3 − 1/4 + ⋯.
36
New cards
What is the value of the alternating harmonic series?
Σ from n=1 to ∞ of (−1)ⁿ⁺¹/n = ln 2.
37
New cards
Why is the alternating harmonic series conditionally convergent?
The original series converges to ln 2, but its absolute-value series is Σ from n=1 to ∞ of 1/n, which diverges.
38
New cards
What should be checked first when determining the character of Σaₙ?
Compute lim aₙ. If aₙ does not tend to 0, the series diverges; if aₙ → 0, further analysis is required.
39
New cards
Which methods should be considered for a series with eventually non-negative terms?
A benchmark series, the direct comparison criterion, or the asymptotic comparison criterion.
40
New cards
What should be done when the signs of a series keep changing?
First study Σ|aₙ|. If it converges, then Σaₙ converges absolutely and therefore converges.
41
New cards
What can be concluded if Σ|aₙ| diverges?
Nothing by itself: Σaₙ may diverge or may converge conditionally.