Proving $a_n=\frac{n}{2}-\lfloor\frac{n}{2}\rfloor$ diverges using $\epsilon-N$

Question

I need to show using only the $\epsilon-N$ definition, that $a_n=\frac{n}{2}-\lfloor\frac{n}{2}\rfloor$ diverges.

So as much as I understand, I need to show that:

For all $L\in\mathbb{R}$, there exisits an $\epsilon$ such that for all $N\in \mathbb{N}$, there exists $n>N$ such that $|a_n-L|\geq \epsilon$.

This seems like total chaos. What epsilon do I choose? how do I express $n$ is terms of $N$?

Answer 1

What is $a_n$? Look carefully, then you will realize: $$ a_n = \begin{cases} \frac{1}2 & n \text{ is odd} \\ 0 & n \text{ is even}\end{cases} $$

Let $L$ be any real number. Then let $p = \min\{|L-0|,|L-\frac{1}{2}|\}$.

Now, if $p > 0$, then choose $\epsilon = \frac{p}{2}$, and note that for all $n \geq 1$, $|a_n - L| \geq p > \epsilon$. That is, given $N \in \mathbb{N}$, we take $M=N+1$, and note that $|a_M - L| > \epsilon$. Hence, the condition is satisfied for these $L$.

Now, suppose that $p=0$. Then $L$ is either $0$ or $\frac{1}{2}$.

Suppose that $L=0$. Note that for odd $m$, $|a_m-L| = \frac{1}{2}$. Just take $\epsilon = \frac{1}{4}$, and note that given $N \in \mathbb{N}$, we take the odd number just after $N$, say $M$, and note that $|a_M-L| = \frac{1}{2} >\epsilon$.

Suppose that $L=\frac{1}{2}$. Note that for even $m$, $|a_m-L| = \frac{1}{2}$. Just take $\epsilon = \frac{1}{4}$, and note that given $N \in \mathbb{N}$, we take the even number just after $N$, say $M$, and note that $|a_M-L| = \frac{1}{2} >\epsilon$.

Hence, for all $L$, we have found an $\epsilon > 0$, such that given $N \in \mathbb{N}$, there exists $M > N$ such that $|a_m - L| > \epsilon$. Hence, $a_n$ diverges.

Answer 2

A good place to start would be to note that $a_n=\frac{1}{2}$ if $n$ is odd, while $a_n=0$ if $n$ is even.

Therefore the only values of $L$ you really need to worry about are $L=0,\frac{1}{2}$. What would be a good choice for $\epsilon$?

Answer 3

If the sequence $(a_n)_{n\geq1}$ were convergent to some $L\in{\mathbb R}$ then there would be an $n_0$ such that $|a_n-L|<{1\over4}$ for all $n>n_0$. The triangle inequality would then imply that $|a_n-a_m|<{1\over2}$ for all $m$, $n>n_0$, but this is obviously not the case.