1. ## Prove by contradiction that the sequence is divergent

This is a tough one, and I have little idea on how to prove it.

Prove that the sequence defined by $f(n)=(-1)^n,n\in N$ is divergent.

This must be proven by contradiction.

2. Originally Posted by Runty
This is a tough one, and I have little idea on how to prove it.

Prove that the sequence defined by $f(n)=(-1)^n,n\in N$ is divergent.

This must be proven by contradiction.

Suppose $(-1)^n\xrightarrow [n\to\infty]{}a$ , for some $a\in\mathbb{R}\Longrightarrow \forall\,\epsilon>0\,\,\,\exists\,N_\epsilon\in\ma thbb{N}\,\,\,s.t.\,\,\,n>N_\epsilon\Longrightarrow \,|(-1)^n-a|<\epsilon$ .

Let us choose $\epsilon=\frac{1}{4}\Longrightarrow$ for some natural number $N_\frac{1}{4}$ , we have that $n>N_\frac{1}{4}\Longrightarrow |(-1)^n-a|<\frac{1}{4}$ $\Longleftrightarrow a-\frac{1}{4}<(-1)^n .

But this means that all but a finite number of the sequence's elements are between the left and the right ends, and the difference between these two ends is $a+\frac{1}{4}-(a-\frac{1}{4})=\frac{1}{2}$ , and this is a contradiction since for any two consecutive natural numbers $n_1,n_2>N_\frac{1}{4}\,,\,\,|(-1)^{n_1}-(-1)^{n_2}|=2$ , so it's impossible that ALL the elements of the seq. (but perhaps a finite number of them) are within two numbers whose difference is $0.5$ ...

Tonio

3. Also, an alternative way would be to notice that if the sequence is convergent then it is cauchy, but for each $n \in \mathbb{N}, \ |a_{n+1} - a_n| =2$

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### to show the sequence is divergent

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