Unbounded sequence diverging to $\infty$?
Andrew Henderson 
If $\{x_n\}$ is a sequence of positive real numbers which is not bounded, then it diverges to infinity. State whether the above statement is true or false. If true/false , give the reason.
I just know that if the sequence is of positive real numbers then it must be either increasing or decreasing sequence or it would be constant sequence.
My question is what can I conclude about the word "not bounded"? Does it means not bounded above or not bounded below or neither bounded above nor bounded below? Anyone just help me to draw conclusion whether the statement is true or not?
$\endgroup$ 63 Answers
$\begingroup$A sequence $(x_{n})$ of reals is bounded iff there is some $M \geq 0$ such that $|x_{n}| \leq M$ for all $n\geq 1$; the sequence $(x_{n})$ diverges to infinity iff for every $B > 0$ we have $|x_{n}| \geq B$ eventually. Now if $(x_{n})$ is not bounded and if $x_{n} > 0$ for all $n \geq 1$, then for every $M > 0$ there is some $n \geq 1$ such that $x_{n} > M$; so $(x_{n})$ need not diverge to infinity by definition. For instance, the sequence $(x_{n})$ defined by $x_{n} := n + (-1)^{n-1}(n-1)$ for all $n \geq 1$ is unbounded and does not diverge to infinity.
$\endgroup$ 3 $\begingroup$The statement is false. For example the sequence 1,2,1,4,1,5,1,6,1,7,... is unbounded sequence of positive real numbers that does not diverge to infinity. It is oscillating sequence.
$\endgroup$ $\begingroup$I just know that if the sequence is of positive real numbers then it must be either increasing or decreasing sequence or it would be constant sequence.
Then what can you say about the sequence defined by
$$a_n = n(1+(-1)^n)$$
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