Question

Prove that the sequence \((a_n)_{n\in \mathbb{n}}\) is bounded iff the sequence \((|a_n|)_{n\in \mathbb{n}}\) is bounded from above.

Answer 1

for one way...
you can suppose |a|<=p for all natural n and p is greater than or equal to 0
|a|<=p implies something about the sequence a

Answer 2

for the other way...
I would use the fact that |x|=sqrt(x^2)

Answer 3

"\(\Rightarrow\)",

Let \(M_1\) be an upper bound for \((a_n)_{n\in \mathbb{N}}\), and \(M_2\) be an lower bound for \((a_n)_{n\in \mathbb{N}}\). Then, we have \(M_2\le a_n \le M_1\) for all n \(\in \mathbb{N}\).
Take \(M = \max(|M_1|, |M_2|)\), then for all n \(\in \mathbb{N}\), we have, \(-M\le M_2\le a_n \le M_1 \le M\), i.e. \(|a_n| \le M\)
Hence, \((|a_n|)_{n\in \mathbb{N}}\) is bounded from above if \((a_n)_{n\in \mathbb{N}}\) is bounded

"\(\Leftarrow\)",

Let \(M\) be an upper bound for \((|a_n|)_{n\in \mathbb{N}}\). Then, for all n \(\in \mathbb{N}\), we have \(|a_n|\le M\), that is \(-M\le a_n\le M\). Hence, \((a_n)_{n\in \mathbb{N}}\) is bounded if \((a_n)_{n\in \mathbb{N}}\) is bounded from above.

Therefore, the sequence \((a_n)_{n\in \mathbb{N}}\) is bounded iff the sequence (\(|a_n|)_{n\in \mathbb{N}}\) is bounded from above.

Does this proof look okay?

Answer 4

looks fine to me
it actually looks more formal than what I have written

Answer 5

We need a formal proof. :)

Answer 6

looks great.