# Thread: Explanation needed on a sequence proof

1. ## Explanation needed on a sequence proof

The Theorem is every convergent sequence is bounded:

$\displaystyle (a_n)\rightarrow\alpha$ Then, by Theorem 2.1 (in my book), $\displaystyle (|a_n|)\rightarrow |\alpha|$.

Quite arbitrary, let us choose $\displaystyle \epsilon =1$ (so I could pick any number not just 1?), and then use the definition of a limit to conclude there exists N such that $\displaystyle \left| |a_n|-|\alpha|\right |< 1, \ \forall n>N$.

That is for n > N,

$\displaystyle |\alpha|-1<|a_n|<|\alpha|+1$.

(I don't understand what is going with the max part)
Hence, $\displaystyle \forall n\geq 1$

$\displaystyle |a_n| \leq \text{max}\{|a_1|,|a_2|,\cdots , |a_N|, |\alpha|+1\}$

and so $\displaystyle (a_n)$ is bounded.

2. Originally Posted by dwsmith
The Theorem is every convergent sequence is bounded: $\displaystyle (a_n)\rightarrow\alpha$ Then, by Theorem 2.1 (in my book), $\displaystyle (|a_n|)\rightarrow |\alpha|$.
choose $\displaystyle \epsilon =1$ , and then use the definition of a limit to conclude there exists N such that $\displaystyle \left| |a_n|-|\alpha|\right |< 1, \ \forall n>N$.
That is for n > N,
$\displaystyle |\alpha|-1<|a_n|<|\alpha|+1$.
(I don't understand what is going with the max part)
Hence, $\displaystyle \forall n\geq 1$
$\displaystyle |a_n| \leq \text{max}\{|a_1|,|a_2|,\cdots , |a_N|, |\alpha|+1\}$ and so $\displaystyle (a_n)$ is bounded.
Actually it is $\displaystyle n\ge N$ then $\displaystyle |a_n|<|\alpha|+1$.
That is, from $\displaystyle N$ on $\displaystyle |\alpha|+1$ bounds the sequence.
But we do not know what happens to $\displaystyle |a_j|$ if $\displaystyle 1\le j<N$.
That is where the max comes in.
Here is the way I taught it: let $\displaystyle M = \sum\limits_{k = 1}^N {\left| {a_k } \right|}$.
Now $\displaystyle \left( {\forall n} \right)$ we have $\displaystyle |a_n|<|\alpha|+1+M$

3. Originally Posted by Plato
Actually it is $\displaystyle n\ge N$ then $\displaystyle |a_n|<|\alpha|+1$.
That is, from $\displaystyle N$ on $\displaystyle |\alpha|+1$ bounds the sequence.
But we do not know what happens to $\displaystyle |a_j|$ if $\displaystyle 1\le j<N$.
That is where the max comes in.
Here is the way I taught it: let $\displaystyle M = \sum\limits_{k = 1}^N {\left| {a_k } \right|}$.
Now $\displaystyle \left( {\forall n} \right)$ we have $\displaystyle |a_n|<|\alpha|+1+M$
Why can you just add M at the end? It is understandable abs(a) + 1 + M is greater than a_n since a_n \leq M so anything added to M would be greater than a_n.

4. Originally Posted by dwsmith
Why can you just add M at the end? It is understandable abs(a) + 1 + M is greater than a_n since a_n \leq M so anything added to M would be greater than a_n.
That is a simple to answer.
If $\displaystyle 1\le n\le N$ then $\displaystyle |a_n|\le M\le M+|\alpha|+1$

If $\displaystyle n\ge N$ then $\displaystyle |a_n|\le |\alpha|+1\le M+|\alpha|+1$