# Thread: limit of an increasing sequence

1. ## limit of an increasing sequence

2. Originally Posted by john_n82

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 $\displaystyle \lim_{n\rightarrow\infty}{x_{n}} = x$

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

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

4. 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 $\displaystyle x_n < x_j$