Equivalence of definition (limited sequence)

Aug 2018
44
0
Denmark
Hello,

I am asked to show that the following two definitions of the limit \(\displaystyle L\) of a sequence \(\displaystyle s_{n}\) are equivalent:

a) \(\displaystyle \forall \epsilon>0 \exists N\in \mathbb{Z} \forall n \geq N: abs(s_{n}-L) < \epsilon\)
b) \(\displaystyle \forall m \in \mathbb{Z}_{+} \exists N\in \mathbb{R} \forall n \geq N: abs(s_{n}-L) < \frac{1}{m}\)

I do see it intuitively, but how do I show it quantitatively?

Thanks.
 
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romsek

MHF Helper
Nov 2013
6,666
3,004
California
$\text{show that $a \Leftrightarrow b$}$
 
Aug 2018
44
0
Denmark
So I must show that:

1) \(\displaystyle a \Rightarrow b\)
2) \(\displaystyle b \Rightarrow a\)

I start with 1):

I assume that a sequence \(\displaystyle \{S_{n}\}\) converges to \(\displaystyle L\) according to 1).
This means that for every \(\displaystyle \epsilon>0\) I can find an \(\displaystyle N\in \mathbb{N}\) such that \(\displaystyle \mid S_{n}-L \mid < \epsilon\) whenever \(\displaystyle n\geq N\).
Since there is no restriction on \(\displaystyle \epsilon\) other than that it has to be positive I let \(\displaystyle \epsilon=m\) where \(\displaystyle m\in \mathbb{Z_{+}}\). Not sure!

Hints would be appreciated!
 
Last edited:
Jun 2013
1,111
590
Lebanon
if we let \(\displaystyle \epsilon=m\) then \(\displaystyle a)\) says

\(\displaystyle \exists N\in \mathbb{Z} \forall n \geq N: abs(s_{n}-L) < m\)

whereas you are trying to show \(\displaystyle < \frac{1}{m}\)
 

Plato

MHF Helper
Aug 2006
22,474
8,643
Hello,

I am asked to show that the following two definitions of the limit \(\displaystyle L\) of a sequence \(\displaystyle s_{n}\) are equivalent:

a) \(\displaystyle \forall \epsilon>0 \exists N\in \mathbb{Z} \forall n \geq N: abs(s_{n}-L) < \varepsilon\)
b) \(\displaystyle \forall m \in \mathbb{Z}_{+} \exists N\in \mathbb{R} \forall n \geq N: abs(s_{n}-L) < \frac{1}{m}\)
I think that you have a typo in part b). Is it not $\Large\exists N\in \mathbb{Z}~?$

So to prove that $a \Rightarrow b$ start with $m \in \mathbb{Z}^{+}$
Using a); suppose that $0<\varepsilon<\frac{1}{m}$ we get $\exists N\in\mathbb{Z}^+$ such that if $n\ge N$ then $|L-s_n|<\varepsilon<\frac{1}{m}$

To prove that $b \Rightarrow a$ start with $\varepsilon>0$.
Using b); we know that $\exists N\in\mathbb{Z}$ having the property that if $n\ge N$ then $|L-s_n|<\varepsilon$
By simple algebra we know that if $n>N $ then $\frac{1}{n}<\frac{1}{N}<\varepsilon$.
Thus if $n\ge N$ then $|L-s_n|<\varepsilon$
 
Aug 2018
44
0
Denmark
Thanks.
The definition states "... for every positive integer m there is a real number N so that ...". I am afraid it is not a typo.
 
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