# Thread: Sequence Bounded Above / Diverge to Infinity

1. ## Sequence Bounded Above / Diverge to Infinity

I'm having a lot of trouble with this one homework problem on sequences. I am given two statements, one is true and one is false. I need to identify the false one and give an example of why it is false.

a) If (a(n)) is not bounded above, then (a(n)) diverges to infinity.
b) If (a(n)) diverges to infinity, then (a(n)) is not bounded above.

Here are the definition we were given word-for-word:
A sequence (a(n)) is bounded above if there exists a U within real numbers such that a(n) <= U for all n within natural numbers.
A sequence (a(n)) diverges to infinity if for all U within real numbers, there exists N within natural numbers so that a(n) >= U for all n >= N.

Right now I'm thinking that (a) is the false one, with a(n) = n*sin(n) as an example. It is not bounded above, but there is no point where a(n) is always greater than or equal to some U, as it is periodically increasing and decreasing, so it does not diverge to infinity. Does this make sense?

2. I suggest using a more clearcut example.

Say $a_n = \left\{ {\begin{array}{rl}
{n,} & {\text{n is even}} \\
{n^{ - 1} ,} & {\text{n is odd}} \\ \end{array} } \right.$

See how this works.