# Equivalence of definition (limited sequence)

#### 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) $$\displaystyle \forall \epsilon>0 \exists N\in \mathbb{Z} \forall n \geq N: abs(s_{n}-L) < \epsilon$$
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
$\text{show that$a \Leftrightarrow b$}$

#### detalosi

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!

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#### Idea

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
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 $|L-s_n|<\varepsilon$

#### detalosi

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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