Show that $\inf(\frac{1}{n})=0$.

Question

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We are given the following definition:

If a sequence $(a_n)$ is bounded from below then there is a greatest lower bound for the sequence called the infimum.

$m=\inf(a_n)$ if

i) $(a_n) \geq m \ \forall n \in \mathbb{N}$.

ii) For each $\epsilon >0 \ \exists \ n_\epsilon \ \in \mathbb{N}$ such that $a_{n_\epsilon} < m + \epsilon $.

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Show using the definition that, if $a_n=\frac{1}{n}$ then $\inf(a_n)=0$.

Here is what I have:

i) RTP: $\frac{1}{n} \geq 0 \ \forall n\in \mathbb{N}$.

Assume that $\frac{1}{n} < 0$ for some $n \in \mathbb{N}$.

$\implies n < 0$.

But we know that $n \in \mathbb{N}, n \geq 1$, and thus we can say that$\frac{1}{n} > 0\ \forall n \in \mathbb{N}$.

ii) Let $ \epsilon > 0$ be given.

RTP: $\exists n_\epsilon \in \mathbb{N}$ such that $\frac{1}{n_\epsilon}<0+\epsilon = \epsilon$.

Proof:

$\frac{1}{n_\epsilon} < \epsilon \implies n_\epsilon > \frac{1}{\epsilon}$, since both values are positive.

For the given $\epsilon > 0$, choose $n_\epsilon > \frac{1}{\epsilon} \in \mathbb{N}$.

Now from the Archimedes Postulate, we know $\exists n \in \mathbb{N}$ such that $n \geq n_\epsilon$.

$\implies n \geq n_\epsilon > \frac{1}{\epsilon}$

$\implies \frac{1}{n} \leq \frac{1}{n_\epsilon} < \epsilon$, since all values are positive.

Answer 1

It's correct but can be written in fewer words:

1. $n>0$ implies $\frac{1}{n}>0$ so $a_n>0$ for all $n>0$;
2. given $\epsilon>0$, just pick any integer $n>\frac{1}{\epsilon}$, then $a_n=\frac{1}{n}<\frac{1}{1/\epsilon}=\epsilon$.

Answer 2

By pure substitution,

$0=\inf\left(\frac1n\right)$ if

i) $\left(\frac1n\right) \geq 0 \ \forall n \in \mathbb{N}$.

ii) For each $\epsilon >0 \ \exists \ n_\epsilon \ \in \mathbb{N}$ such that $\frac1{n_\epsilon} < \epsilon $.

Then

i) is obvious.

ii) holds by taking $n_\epsilon=\lceil\frac1\epsilon\rceil$.