2
$\begingroup$

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:

  1. Could you please tell me if my proof is right and how and where I could improve it?

  2. Is there a better / more elegant way to prove this?

It would be nice if any arguments were as rigorous as possible.

$\endgroup$
2
  • $\begingroup$ $\forall$ C $>$ 0, there exists a N $\in$ the naturals such that $\forall$ n $>$ N, $(a_n)$ $>$ C @Piwi $\endgroup$
    – Robbie
    Commented Oct 11, 2015 at 22:16
  • $\begingroup$ (it may be a silly question) so can we assume that $a_n > a_k\,\,\forall n>k$ $\endgroup$
    – Chinny84
    Commented Oct 12, 2015 at 18:09

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Seems a tad bit vague, can you or anyone please write a solid proof? $\endgroup$
    – Robbie
    Commented Oct 11, 2015 at 22:18
  • $\begingroup$ @Robbie Which part is it you don't understand or feels vague to you? $\endgroup$
    – Piwi
    Commented Oct 11, 2015 at 22:55
  • $\begingroup$ @Piwi, This is what I wrote Let C $>$ 0 be arbitrary Let $(a_n)$ -> $\infty$ as n -> $\infty$ By definition, $\forall$ C $>$ 0, there exists $N_1$ $\in$ the Naturals such that n $>$ $N_1$, $a_n$ $>$ C That indicates that there exists $N_1$ $\in$ the Naturals such that $n+2 > n+1 > n > N_1$ This imples that ... $\geq$ $a_n+2$ $\geq$ $a_n+1$ $\geq$ $a_n$ $>$ C. So by definition of a increasing sequence, $(a_n)$ is increasing as it tends to infinity. This indicates that there doesn't exist u $\in$ $\Re$ s.t. $\forall$ n $\in$ the Naturals, u $\geq$ $a_n$ because... $\endgroup$
    – Robbie
    Commented Oct 12, 2015 at 0:05
  • $\begingroup$ $\forall$ u $\in$ $\Re$ there exists a $N_2$ $\in$ the Naturals s.t. $\forall$ n $>$ $N_2$, $a_n$ $>$ u Let $N$ = $max(N_1, N_2)$ Then $\forall$ n $>$ $N$, $a_n$ $>$ u $\endgroup$
    – Robbie
    Commented Oct 12, 2015 at 0:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .