1. Absolute Convergence Ratio Test Root Test
Lesson 11-6
Absolute Convergence Ratio Test Root Test

2. Vocabulary
Absolute Convergent – a series of numbers that alternate in sign, like the summation of the following
Conditionally Convergent -- a convergent series that is not absolutely convergent
Rearrangement – a reordering of terms in an infinite series

3. Types of Series
Geometric
Telescoping
Harmonic
P-Series
Alternating

4. Types of Series Tests Divergence (easiest to check)
if lim an ≠ 0 then it diverges
if lim an = 0 then it is inconclusive
Comparison (last resort)
if an ≤ bn for all n 0 then it diverges
Limit Comparison (better than comparison)
Ratio (apply when factorials are involved)
Root (apply when nth power involved)
n→∞

5. Ratio Test | | | | Let ∑ an be a series
If , then the series ∑an is absolutely convergent (and therefore convergent)
If , or , then the series ∑an is divergent
If , the Ratio test is inconclusive (no conclusion about the convergence or divergence of the series can be drawn)
an+1
lim = L < 1 an |
n→∞
an+1
lim = L > 1 an |
n→∞
an+1
lim = ∞ an
n→∞ |
an+1
lim = 1 an
n→∞ |

6. Root Test Let ∑ an be a series
If , then the series ∑an is absolutely convergent (and therefore convergent)
If , or , then the series ∑an is divergent
If , the Ratio test is inconclusive (no conclusion about the convergence or divergence of the series can be drawn)
lim = L < 1
n→∞
√|an| n
lim = L > 1
n→∞
√|an| n
lim = ∞
n→∞
√|an| n
lim = 1
n→∞
√|an| n

7. 11-6 Example 1 | Is the following series convergent or divergent?
(-1)n+1 2n
series n!
Lim bn = indeterminate and L’Hospital’s Rule does not apply!
n→∞
Use Ratio Test, the absolute values eliminate the (-1) term and we get
an n+1 / (n+1)!
lim = lim
an n / n! |
n→∞
n! 2n
This leads to lim = = 0 therefore it converges
(n+1)! 2n n+1 7

8. 11-6 Example 2 Is the following series convergent or divergent? 6n
Lim = indeterminate and L’Hospital’s Rule does not apply!
n→∞
6n n - 1 n
Ratio test is impossible, so we can try the Root test
Lim = Lim = 2
n→∞
6n n - 1 n
1/n
since p > 1 the series diverges by the Root Test 8

9. Homework
Pg : problems 3, 4, 7, 11