# Showing the sequence is bounded.

#### simpleas123

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! #### Plato

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

#### simpleas123

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?

#### Plato

MHF Helper
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?

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• simpleas123

#### simpleas123

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!

#### Bruno J.

MHF Hall of Honor
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.

#### mabruka

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.

• Defunkt and Plato