# Thread: Equivalence of definition (limited sequence)

1. ## Equivalence of definition (limited sequence)

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.

2. ## Re: Equivalence of definition (limited sequence)

$\text{show that$a \Leftrightarrow b$}$

3. ## Re: Equivalence of definition (limited sequence)

So I must show that:

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

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!

4. ## Re: Equivalence of definition (limited sequence)

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 $\displaystyle < \frac{1}{m}$

5. ## Re: Equivalence of definition (limited sequence)

Originally Posted by detalosi
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 $|L-s_n|<\varepsilon$

6. ## Re: Equivalence of definition (limited sequence)

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.