limit and supremum

**Zeke** · January 17th 2012, 02:13 AM

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 $x_n<x_{n+1}$ $\forall n$ and $lim_{n\rightarrow \infty}(x_n/x_{n+1})=1$ . Can I then, knowing the above, conclude that $\sup\{x_n/x_{n+1}; n\in\mathbb{N}\}=1$ ?

**Plato** · January 17th 2012, 02:32 AM

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 $x_n<x_{n+1}$ $\forall n$ 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 $0<a<b$ then is it true that $\frac{a}{b}<1~?$

**Zeke** · January 17th 2012, 04:08 AM

Yes, ofcourse it's true.

**Plato** · January 17th 2012, 06:51 AM

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

**psolaki** · March 3rd 2012, 05:00 PM

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

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