# Explanation needed on a sequence proof

• Jun 2nd 2011, 12:44 PM
dwsmith
Explanation needed on a sequence proof
The Theorem is every convergent sequence is bounded:

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

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

That is for n > N,

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

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

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

and so $(a_n)$ is bounded.
• Jun 2nd 2011, 01:10 PM
Plato
Quote:

Originally Posted by dwsmith
The Theorem is every convergent sequence is bounded: $(a_n)\rightarrow\alpha$ Then, by Theorem 2.1 (in my book), $(|a_n|)\rightarrow |\alpha|$.
choose $\epsilon =1$ , and then use the definition of a limit to conclude there exists N such that $\left| |a_n|-|\alpha|\right |< 1, \ \forall n>N$.
That is for n > N,
$|\alpha|-1<|a_n|<|\alpha|+1$.
(I don't understand what is going with the max part)
Hence, $\forall n\geq 1$
$|a_n| \leq \text{max}\{|a_1|,|a_2|,\cdots , |a_N|, |\alpha|+1\}$ and so $(a_n)$ is bounded.

Actually it is $n\ge N$ then $|a_n|<|\alpha|+1$.
That is, from $N$ on $|\alpha|+1$ bounds the sequence.
But we do not know what happens to $|a_j|$ if $1\le j.
That is where the max comes in.
Here is the way I taught it: let $M = \sum\limits_{k = 1}^N {\left| {a_k } \right|}$.
Now $\left( {\forall n} \right)$ we have $|a_n|<|\alpha|+1+M$
• Jun 2nd 2011, 01:22 PM
dwsmith
Quote:

Originally Posted by Plato
Actually it is $n\ge N$ then $|a_n|<|\alpha|+1$.
That is, from $N$ on $|\alpha|+1$ bounds the sequence.
But we do not know what happens to $|a_j|$ if $1\le j.
That is where the max comes in.
Here is the way I taught it: let $M = \sum\limits_{k = 1}^N {\left| {a_k } \right|}$.
Now $\left( {\forall n} \right)$ we have $|a_n|<|\alpha|+1+M$

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.
• Jun 2nd 2011, 01:30 PM
Plato
Quote:

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

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