Problem:
Prove that $(a_n)$ -> $\infty$ implies that $(a_n)$ is not bounded above.
My attempt:
Let $C>0$ be arbitrary. Let $(a_n)\to\infty$ as $n\to\infty$.
By definition, $\forall C>0$, there exists $N_1\in\mathbb{N}$ such that $$\forall n>N_1: a_n>C.$$
That indicates that there exists $N_1\in\mathbb{N}$ such that $$n+2>n+1>n>N_1.$$
This implies that $$\ldots \geq a_n+2\geq a_n+1\geq a_n>C.$$
So by the definition of an increasing sequence, $(a_n)$ is increasing as it tends to infinity. This indicates that there doesn't exist $u\in\mathbb{R}$ s.t. $$\forall n\in\mathbb{N}: u\geq a_n$$
because for every $u\in\mathbb{R}$ there exists a $N_2\in\mathbb{N}$ s.t. $\forall n>N_2: a_n>u$.
Let $N=\max(N_1,N_2)$. Then $\forall n>N: a_n>u$.
My questions:
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Could you please tell me if my proof is right and how and where I could improve it?
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Is there a better / more elegant way to prove this?
It would be nice if any arguments were as rigorous as possible.
Best Answer
Let $(a_n) \to \infty$. Assumption: $(a_n)$ is bounded above.
Therefore, there exists $u \in \mathbb{R}_{>0}$ such that $$\forall n \in \mathbb{N}: a_n\leq u.$$
But in the beginning we let $(a_n) \to \infty$ which according to your definition means that for all $C>0$, especially for $C:=u>0$, there exists $N\in\mathbb{N}$ such that $$\forall n> N:a_n>u.$$ As the two statements above contradict each other, our assumption must be wrong and so $(a_n)$ is not bounded above.