- #1

- 1

- 0

2. If {an} and {bn} are convergent and there exists a constant k>0 such that |bn| > k for all n=1, 2, ...., then {an/bn} is also convergen

- Thread starter darthprince
- Start date

- #1

- 1

- 0

2. If {an} and {bn} are convergent and there exists a constant k>0 such that |bn| > k for all n=1, 2, ...., then {an/bn} is also convergen

- #2

CompuChip

Science Advisor

Homework Helper

- 4,302

- 47

E.g. you can try to show that

[tex]\forall_{\epsilon > 0} \exists_{N_+} : n > N_+ \implies |(a_n + b_n) - L| < \epsilon[/tex]

where L = a + b is the postulated limit. Of course you already know that given such an [itex]\epsilon[/itex] you can find N

[tex]n > N_a \implies |a_n - a| < \epsilon[/tex]

and

[tex]n > N_b \implies |b_n - b| < \epsilon[/tex]

- Last Post

- Replies
- 13

- Views
- 9K

- Replies
- 4

- Views
- 10K

- Replies
- 3

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 7K

- Replies
- 4

- Views
- 2K

- Replies
- 3

- Views
- 11K

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 3K

- Replies
- 15

- Views
- 7K