# limit of an increasing sequence

• Apr 15th 2009, 08:44 PM
john_n82
limit of an increasing sequence
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• Apr 16th 2009, 04:00 AM
xalk
Quote:

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

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

GIVEN ε>0 there exists kεN AND SUCH that:

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

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

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

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

hence :

$x-\epsilon

THUS $\lim_{n\rightarrow\infty}{x_{n}} = x$
• Apr 16th 2009, 09:56 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 $n\geq k$ $x_{n}\geq x_{k}$

• Apr 16th 2009, 10:03 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, $x_n < x_{n + 1} < x_{n + 2} < \cdots < x_{n + K}$.
So if $j>n$ then $x_n < x_j$