Thread: 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

$\displaystyle
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?

Thanks for your help!

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

Originally Posted by Plato

What am I missing?
$\displaystyle \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?
And i dont get what your asking me?

Originally Posted by simpleas123

Why did u just take the absolute value of an?
And i dont get what your asking me?

I don't get what it is that you do not understand.
Do you understand what it means to be bounded?

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!
$\displaystyle
| a_n | \leqslant M
$
For some M.
In this case, being that an converges to 1 therefore M = 1.

Thank you very much!

Originally Posted by simpleas123

Ahh, its so simple!
$\displaystyle
| a_n | \leqslant M
$
For some M.
In this case, being that an converges to 1 therefore M = 1.

Thank you very much!

$\displaystyle a_n$ does not converge to 1.

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


Boundedness is a concept that comes before convergence,


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

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


Convergence has nothing to do with boundedness so far.


Thats why Plato did what he did.