# Convergence question

1. Dec 3, 2008

### transgalactic

prove or desprove that if An->1

then An+1/An ->1

??

2. Dec 3, 2008

### Office_Shredder

Staff Emeritus
Anytime the question says "prove or disprove" and you're not sure what to do, start by spending five minutes trying to generate a counterexample. If you can't do it, see if you can figure out what's stopping you (which usually leads to a proof that the statement is true)

3. Dec 3, 2008

### transgalactic

i know that if a series is converges and monotone
then if An->1 then An+1 ->1

so i construct a limit An+1/An
n-> +infinity
and say that if both members go to 1 then the whole expression goes to 1

the problem is:
i dont know if An is monotone
and if this way is correct and will be regarded as a formal proof
??

4. Dec 3, 2008

### Office_Shredder

Staff Emeritus
You don't need the monotone condition here. For all e>0, there exists N>0 such that n>N implies |An - 1| < e gives us that |An+1 - 1| < e for n>N also as n+1>n>N

5. Dec 3, 2008

### transgalactic

and after writing what you said
i do the limit part?

6. Dec 3, 2008

### Office_Shredder

Staff Emeritus
Yeah

7. Dec 4, 2008

thanks

8. Dec 4, 2008

### HallsofIvy

Staff Emeritus
You have been posting questions here long enough to have learned to copy the problem correctly even if you do not wish to use Latex.

Is this A_{n+1}/A_n or (A_n+ 1)/A_n ?

9. Dec 4, 2008

### Staff: Mentor

Or maybe it's $$A_n + \frac{1}{A_n}$$

10. Dec 4, 2008

### transgalactic

its A(n+1)/A(n)

11. Dec 5, 2008

### transgalactic

when i solve this type of question:
how do i recognize that the expression doesnt converge
and how do i disprove that the function converges
in such case?

12. Dec 5, 2008

### transgalactic

i need to prove that (An)^n ->1

but when i construct limit
lim (An)^n
n->+infinity

i get 1^(+infinity) which says that there is no limit
what do i do in this case in order to disprove that (An)^n->1

??

13. Dec 5, 2008

### HallsofIvy

Staff Emeritus
Why do you need to prove that Ann goes to 1? Is this a completely different question?

As to your original question, if An converges to 1, then, given any $\epsilon> 0$ there exist N such that if n> N, $1-\epsilon\le A_n\le 1+\epsilon$. If n> N, that's true for both An and An+1.