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Prove that if sequence is unbounded, it diverges.

We start by recounting the definition of an unbounded sequence.

  1. Unbounded sequence : A sequence is said to be unbounded if $\forall k \in \mathbb{R}, \exists m \in \mathbb{N} : \lvert a_m \rvert≥k$.

We Proceed by contradiction. That is, let the sequence converge.

We now recall the definition of convergence.

  1. A sequence $(a_n)$ is said to converge to $L$iff $\forall \varepsilon>0, \exists N \in \mathbb{N}: n≥N \implies \lvert a_n - L \rvert< \varepsilon$

There are two cases. The $m$ such that $\forall k \in \mathbb{R}, \exists m \in \mathbb{N} : \lvert a_m\rvert≥k$ is more than or equal to the $N$ such $ \forall \varepsilon>0 ,\lvert a_n-L\rvert<\varepsilon$. Or $m$ is less than this $N$.

Case 1: The m is greater than the $N$

Here we have $a_n<\varepsilon+L$, where $\varepsilon+L$ is an arbitrary real number. At the same time we have $a_n>k$ where k is an arbitrary real number. This is a contradiction. Thus our assumption that $a_n$ converges is wrong.

Case 2: The idea here is to show this is false. Unsure of how to do that, would like a hint.

math and physics forever
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1 Answers1

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You should be careful with the definition of limit. There is another quantified variable implicitly present. A correct definition of limit would be $$\lim a_n = L \Leftrightarrow (\forall\varepsilon >0)(\exists N\in\mathbb N)(\forall n\in\mathbb N)(n>N \Rightarrow |a_n-L|<\varepsilon).$$

Regarding a direct proof.

Let $(a_n)$ be unbounded and fix arbitrary $L\in\mathbb R$. We want to show that $\lim a_n \neq L$. That is, we want to find an $\varepsilon _0>0$ such that $$(\forall n\in\mathbb N)(\exists m = m(n)\in\mathbb N)(m>n\quad\&\quad |a_m-L|\geqslant\varepsilon_0).$$ Take $\varepsilon _0 = 1$ and let $n\in\mathbb N$. By unboundedness there exists an index $m>n$ such that $|a_m| > |L|+1$. Conclude $$1<|a_m|-|L|\leqslant |\,|a_m|-|L|\,| \leqslant |a_m-L|.$$ Thus an unbounded sequence cannot have a finite limit.

AlvinL
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