Is the Product of a Null Sequence and a Bounded Sequence Always Null?

[phospho]

Prove that if (a_n) is a null sequence and (b_n) is a bounded sequence then the sequence (a_nb_n) is null:

from definitions if b_n is bounded then $\exists H \in \mathbb{R}$ s.t. $|b_n| \leq H$ if a_n is a null sequence it converges to 0 (from my book), i.e. given $\epsilon ' > 0$ $\exists N \in \mathbb{R}$ s.t. n>N $\Rightarrow |a_n| < \epsilon '$ set $e' = \dfrac{\epsilon}{|H|}$ then for $n > N \Rightarrow |a_nb_n | < \epsilon ' |H| = \epsilon$ i.e. a_nb_n also converges to 0.

I'm slightly worried about this, if H is negative, then can we just multiply $|a_n|$ and $|b_n|$ as afaik from the axioms if $a>0$ $x>y$ then $xa>ya$ and I'm not sure if H is negative or if |a_n| is negative if we can simply just multiply the two inequalities. Any help, explaining this, thanks.

 

[LCKurtz]

Prove that if (a_n) is a null sequence and (b_n) is a bounded sequence then the sequence (a_nb_n) is null:

from definitions if b_n is bounded then $\exists H \in \mathbb{R}$ s.t. $|b_n| \leq H$ if a_n is a null sequence it converges to 0 (from my book), i.e. given $\epsilon ' > 0$ $\exists N \in \mathbb{R}$ s.t. n>N $\Rightarrow |a_n| < \epsilon '$ set $e' = \dfrac{\epsilon}{|H|}$ then for $n > N \Rightarrow |a_nb_n | < \epsilon ' |H| = \epsilon$ i.e. a_nb_n also converges to 0.

I'm slightly worried about this, if H is negative, then can we just multiply $|a_n|$ and $|b_n|$ as afaik from the axioms if $a>0$ $x>y$ then $xa>ya$ and I'm not sure if H is negative or if |a_n| is negative if we can simply just multiply the two inequalities. Any help, explaining this, thanks.

How can $|a_n|$ be negative?

 

[LCKurtz]

Prove that if (a_n) is a null sequence and (b_n) is a bounded sequence then the sequence (a_nb_n) is null:

from definitions if b_n is bounded then $\exists H \in \mathbb{R}$ s.t. $|b_n| \leq H$ if a_n is a null sequence it converges to 0 (from my book), i.e. given $\epsilon ' > 0$ $\exists N \in \mathbb{R}$ s.t. n>N $\Rightarrow |a_n| < \epsilon '$ set $e' = \dfrac{\epsilon}{|H|}$ then for $n > N \Rightarrow |a_nb_n | < \epsilon ' |H| = \epsilon$ i.e. a_nb_n also converges to 0.

Since your proof is correct, I think I can get away with showing you how to write it up a bit nicer:

Since $\{b_n\}$ is bounded there is $H \in \mathbb{R}$ such that $|b_n|\le H$ for all natural numbers $n$. Suppose $\epsilon > 0$. Since $a_n\to 0$ there exists $N\in \mathbb R$ such that if $n>N$ we have $|a_n| < \frac \epsilon H$. Now if $n>N$ then we have $|a_nb_n-0| = |a_n||b_n| < \frac \epsilon H\cdot H =\epsilon$. Hence $a_nb_n\to 0$.

 

[phospho]

How can $|a_n|$ be negative?

ok $|a_n| < \epsilon$ this is the distance from a_n to 0 which is less than epsilon so I guess it can't be negative. But what if $|b_n| \leq H$ and H is negative? take e.g. $|b_n| \leq -1$ then $1 \leq b_n \leq -1$

 

[LCKurtz]

take e.g. $|b_n| \leq -1$ then $1 \leq b_n \leq -1$

Doesn't $1\le -1$ bother you? Or $|b_n|\le -1$?

 

[phospho]

Doesn't $1\le -1$ bother you? Or $|b_n|\le -1$?

ha - yeah, thanks.

Also thanks for writing up a more concise proof!