Series / Tests

Question 1

What is the P-Series and what determines its Convergence/Divergence

Answer 1

Any Series in the form of: 1/nᵖ

if P > 1 = Convergence
if P ≤ 1 = Diverges

Question 2

What form is a Geometric Series & what is the formula to find it’s sum ?

Answer 2

Σa(R)ᵏ

Defined upper bound: Sₖ = Σ a(1 - Rᵏ⁻¹)/(1 - R)

Infinite Series: S = Σ a/(1 - R)

Question 3

What are the Conditions of the Comparative Test ?

Answer 3

Aₙ ≤ Bₙ

if ΣBₙ converges then ΣAₙ converges
if ΣAₙ diverges then ΣBₙ diverges

Question 4

How is the Limit Comparison Test Performed ?, What are the conditions ?

Answer 4

By reducing the Series to one that appears simpler in nature but has a similar tendency to the original Series. Then divide the two series by each other and take the limit at n approaches ∞

if 0 < L < ∞, our functions act the same

if L = 0 AND simpler series (ΣBₖ) converges, our original (ΣAₖ) Converges too

if L = ∞ AND simpler series (ΣBₖ) diverges, our original (ΣAₖdiverges too

Question 5

When and how is the Alternating Series test performed ?

Answer 5

When the series looks like: Σ (-1)ⁿbₙ

It is convergent if:

limit n → ∞ of bₙ = 0
{bₙ} is a decreasing sequence

Question 6

How is this series solved ?

Answer 6

Take the orignal value and divide it buy 1- r:

(1/3⁴)/(1 - 1/3⁴) = 1/80

Question 7

What is the nth term of this telescoping series ?

How do we solve to the S sum of the sequence ?

Answer 7

The nth term is found by solving the series until we notice a canceling pattern:

(5/1 - 5/2) + (5/2 - 5/3) + (5/3 - 5/4) +—–

we see that the first of each part of the series is the only part that is kept so we retain it in the series and replace any k with n value:

Sₙ = [5 - 5/(n + 1)]

We then solve the series by taking the limit n → ∞

S = 5