# Thread: Sequence, Increasing, Not Bounded

1. ## Sequence, Increasing, Not Bounded

Hello,

Can someone explain to me why this is considered not bounded?

After looking up the definition. I can't seem to understand why this wouldn't be considered bounded.

If there exists a number m such that for every n we say the sequence is bounded below. The number m is sometimes called a lower bound for the sequence.

If there exists a number M such that for every n we say the sequence is bounded above. The number M is sometimes called an upper bound for the sequence.

2. ## Re: Sequence, Increasing, Not Bounded

Originally Posted by l flipboi l
Hello,

Can someone explain to me why this is considered not bounded?

After looking up the definition. I can't seem to understand why this wouldn't be considered bounded.
Put:

$\displaystyle a_n=3n+\frac{1}{n}$

Suppose it is bounded above by $\displaystyle N >0$ which we may suppose an integer, then

$\displaystyle a_n=3n+\frac{1}{n}<N$

Can you see why this is impossible?

Hint $\displaystyle a_n>3n$ so consider $\displaystyle a_N$.

CB

3. ## Re: Sequence, Increasing, Not Bounded

Thanks! I think I get it.

4. ## Re: Sequence, Increasing, Not Bounded

okay, I think I know why. Please correct me if i'm wrong, but the reason why this is not bounded is because there is no limit. It grows as the sequence increases.

5. ## Re: Sequence, Increasing, Not Bounded

it would be better to say it is unbounded, because there is no bound (of course, this also means there is no limit...well, i almost hate to say this, because some people allow "infinte limits" as a valid way of expressing unbounded sequences). as CaptainBlack pointed out, assuming there could BE a bound, leads to a contradiction.

the idea is, if there is a bound, this bound is some finite number we could (in theory, anyway) actually find. but if no matter what finite number we pick, we can find a term that exceeds that bound, there must not be a bound in the first place (the term number that exceeds a given bound could conceivably be, for some sequences, quite a bit larger than the bound itself. fortunately, in this example, we can use the same "N" term as the bound we try).

Thank you!