Thread: Showing the sequence is bounded.

1. Showing the sequence is bounded.

Hey all i thought i understood this question, but after having another look i think my solution is wrong.

"Show that the sequence {an} is bounded, where an, is given by

$
a_n = \frac {(-1)^n}{n^3}
$
"

I know what if i show {an} is convergent, then it is therefore bounded.

My initial idea was to show that an was convergent by the alternating series test.

But thats the problem, its not a series.
So is my method correct, or shall i be approaching this differently?

2. What am I missing?
$\left| {\frac{{\left( { - 1} \right)^n }}
{{n^3 }}} \right| = \frac{1}
{{n^3 }} \leqslant 1
$

3. Originally Posted by Plato
What am I missing?
$\left| {\frac{{\left( { - 1} \right)^n }}
{{n^3 }}} \right| = \frac{1}
{{n^3 }} \leqslant 1
$
Why did u just take the absolute value of an?

4. Originally Posted by simpleas123
Why did u just take the absolute value of an?
I don't get what it is that you do not understand.
Do you understand what it means to be bounded?

5. Originally Posted by Plato
I don't get what it is that you do not understand.
Do you understand what it means to be bounded?
Ahh, its so simple!
$
| a_n | \leqslant M
$

For some M.
In this case, being that an converges to 1 therefore M = 1.

Thank you very much!

6. Originally Posted by simpleas123
Ahh, its so simple!
$
| a_n | \leqslant M
$

For some M.
In this case, being that an converges to 1 therefore M = 1.

Thank you very much!
$a_n$ does not converge to 1.

7. Leave alone for a moment the statement: "If $(a_n)_n$ is convergent then it is bounded".

Boundedness is a concept that comes before convergence,

"A sequence $(a_n)_n$ is bounded if there is a number M such that

$|a_n|\leq M$ for all n.

Convergence has nothing to do with boundedness so far.

Thats why Plato did what he did.