# limit of an increasing sequence

• Apr 15th 2009, 07:44 PM
john_n82
limit of an increasing sequence
http://img16.imageshack.us/img16/8976/question6.jpg
• Apr 16th 2009, 03:00 AM
xalk
Quote:

Originally Posted by john_n82
http://img16.imageshack.us/img16/8976/question6.jpg

Since x is the supremum of {$\displaystyle x_{n}:n\in n$},

GIVEN ε>0 there exists kεN AND SUCH that:

$\displaystyle x-\epsilon< x_{k}\leq x<x+\epsilon$

But since {$\displaystyle x_{n}$} is increasing and bounded by x:

for all n $\displaystyle x_{n}\leq x$ and

for $\displaystyle n\geq k$ $\displaystyle x_{n}\geq x_{k}$

hence :

$\displaystyle x-\epsilon<x_{n}\leq x<x+\epsilon\Longleftrightarrow|x_{n}-x|<\epsilon$

THUS $\displaystyle \lim_{n\rightarrow\infty}{x_{n}} = x$
• Apr 16th 2009, 08:56 AM
john_n82
Thank you for responding. There is a line that I don't understand. Could you please explain further more?

Quote:

Originally Posted by xalk
for $\displaystyle n\geq k$ $\displaystyle x_{n}\geq x_{k}$

• Apr 16th 2009, 09:03 AM
Plato
Quote:

Originally Posted by john_n82
Thank you for responding. There is a line that I don't understand. Could you please explain further more?

You know that the sequence is increasing.

Therefore, $\displaystyle x_n < x_{n + 1} < x_{n + 2} < \cdots < x_{n + K}$.
So if $\displaystyle j>n$ then $\displaystyle x_n < x_j$