1. limit and supremum

Ok, this might be a silly/easy question but I have to make sure. Let say we have a seq of pos. numbers $\{x_n\}$ such that $x_n $\forall n$ and $lim_{n\rightarrow \infty}(x_n/x_{n+1})=1$. Can I then, knowing the above, conclude that $\sup\{x_n/x_{n+1}; n\in\mathbb{N}\}=1$?

2. Re: limit and supremum

Originally Posted by Zeke
Ok, this might be a silly/easy question but I have to make sure. Let say we have a seq of pos. numbers $\{x_n\}$ such that $x_n $\forall n$ and $lim_{n\rightarrow \infty}(x_n/x_{n+1})=1$. Can I then, knowing the above, conclude that $\sup\{x_n/x_{n+1}; n\in\mathbb{N}\}=1$?
If $0 then is it true that $\frac{a}{b}<1~?$

3. Re: limit and supremum

Yes, ofcourse it's true.

4. Re: limit and supremum

Originally Posted by Zeke
Yes, ofcourse it's true.
Well then each term of $\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\} < 1$.
Thus $\sup \left( {\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\}} \right) \le 1$.

What is the contradiction if we suppose that $\sup \left( {\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\}} \right)<1~?$

5. Re: limit and supremum

Originally Posted by Plato
Well then each term of $\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\} < 1$.
Thus $\sup \left( {\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\}} \right) \le 1$.

What is the contradiction if we suppose that $\sup \left( {\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\}} \right)<1~?$

$\sup \left( {\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\}} \right)<\frac{a_{n}}{a_{n+1}}$

For some n