# limit and supremum

• Jan 17th 2012, 02:13 AM
Zeke
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 $\displaystyle \{x_n\}$ such that $\displaystyle x_n<x_{n+1}$ $\displaystyle \forall n$ and $\displaystyle lim_{n\rightarrow \infty}(x_n/x_{n+1})=1$. Can I then, knowing the above, conclude that $\displaystyle \sup\{x_n/x_{n+1}; n\in\mathbb{N}\}=1$? (Thinking)
• Jan 17th 2012, 02:32 AM
Plato
Re: limit and supremum
Quote:

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

If $\displaystyle 0<a<b$ then is it true that $\displaystyle \frac{a}{b}<1~?$
• Jan 17th 2012, 04:08 AM
Zeke
Re: limit and supremum
Yes, ofcourse it's true.
• Jan 17th 2012, 06:51 AM
Plato
Re: limit and supremum
Quote:

Originally Posted by Zeke
Yes, ofcourse it's true.

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

What is the contradiction if we suppose that $\displaystyle \sup \left( {\left\{ {\frac{{a_n }}{{a_{n + 1} }}} \right\}} \right)<1~?$
• Mar 3rd 2012, 05:00 PM
psolaki
Re: limit and supremum
Quote:

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

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

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

For some n