Proving the Convergence of An->1: A Debate on An+1/An Ratio

[transgalactic]

prove or desprove that if An->1


then An+1/An ->1


??

 

[Office_Shredder]

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)

 

[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 don't know if An is monotone
and if this way is correct and will be regarded as a formal proof
??

 

[Office_Shredder]

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

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

 

[transgalactic]

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

 

[Office_Shredder]

Yeah

 

[transgalactic]

thanks

 

[HallsofIvy]

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 ?

 

[Mark44]

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

 

[transgalactic]

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

 

[transgalactic]

when i solve this type of question:
how do i recognize that the expression doesn't converge
and how do i disprove that the function converges
in such case?

 

[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

??

 

[HallsofIvy]

Why do you need to prove that A<sub>nⁿ goes to 1? Is this a completely different question?

As to your original question, if An converges to 1, then, given any \epsilon&gt; 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.</sub>