# sum, difference and product of two null-sequence (need some help)

• Jul 20th 2010, 05:46 AM
yeKciM
[SOLVED] sum, difference and product of two null-sequence (need some help)
It's like this. (Thinking)
I didn't have any problems to prove some of definitions of sequences like :

$\displaystyle lim_{n->+\infty}{(ax_n+by_n)}=ax+by$

Let
$\displaystyle lim_{n->+\infty}{x_n}=x$
and
$\displaystyle lim_{n->+\infty}{y_n}=y$
for any $\displaystyle \xi>0$. Starting from some index $\displaystyle n_1$ every member of sequence $\displaystyle (x_n)$ are in $\displaystyle \xi$-region of point x. Same, from some index $\displaystyle n_2$ every member of sequence $\displaystyle (y_n)$ are in $\displaystyle \xi$-region of point y.

If we put $\displaystyle n_0=max {\{n_1 ,n_2 \}}$ then for every $\displaystyle n >= n_0$ these inequality are true :

$\displaystyle |x_n-x|<\xi$ and $\displaystyle |y_n-y|<\xi$

so from inequality of triangle for $\displaystyle n >= n_0$ we have:

$\displaystyle |ax_n+by_n-(ax+by) |=|ax_n-ax+by_n-by|\le |a||x_n-x|+|b||y_n-y| \le (|a|+|b| )\xi$

Because $\displaystyle (|a|+|b|)$ is fix real number and $\displaystyle \xi$ any small number so it's $\displaystyle (|a|+|b| )\xi$ any small number so there we have that is:

$\displaystyle lim_{n->+\infty}{(ax_n+by_n)}=ax+by$

true.

This and some another I got it OK (Wink)

But I can't, (Headbang)(Headbang)(Headbang) or don't see how to prove (or show) that sum of two null-sequences are again null-sequence (and difference of two sequences) becose it's to obviously. And for the product of two null-sequences is null-sequence and how we can relax conditions for the product of two null-sequences?

any help will be most appreciated (Thinking)
Thanks very much (Hi)(Bow)
• Jul 20th 2010, 11:03 AM
Failure
Quote:

Originally Posted by yeKciM
It's like this. (Thinking)
I didn't have any problems to prove some of definitions of sequences like :

$\displaystyle lim_{n->+\infty}{(ax_n+by_n)}=ax+by$

Let
$\displaystyle lim_{n->+\infty}{x_n}=x$
and
$\displaystyle lim_{n->+\infty}{y_n}=y$
for any $\displaystyle \xi>0$. Starting from some index $\displaystyle n_1$ every member of sequence $\displaystyle (x_n)$ are in $\displaystyle \xi$-region of point x. Same, from some index $\displaystyle n_2$ every member of sequence $\displaystyle (y_n)$ are in $\displaystyle \xi$-region of point y.

If we put $\displaystyle n_0=max {\{n_1 ,n_2 \}}$ then for every $\displaystyle n >= n_0$ these inequality are true :

$\displaystyle |x_n-x|<\xi$ and $\displaystyle |y_n-y|<\xi$

so from inequality of triangle for $\displaystyle n >= n_0$ we have:

$\displaystyle |ax_n+by_n-(ax+by) |=|ax_n-ax+by_n-by|\le |a||x_n-x|+|b||y_n-y| \le (|a|+|b| )\xi$

Because $\displaystyle (|a|+|b|)$ is fix real number and $\displaystyle \xi$ any small number so it's $\displaystyle (|a|+|b| )\xi$ any small number so there we have that is:

$\displaystyle lim_{n->+\infty}{(ax_n+by_n)}=ax+by$

true.

This and some another I got it OK (Wink)

But I can't, (Headbang)(Headbang)(Headbang) or don't see how to prove (or show) that sum of two null-sequences are again null-sequence (and difference of two sequences) becose it's to obviously.

Well, that's funny: you just wrote that you have no trouble believing (even proving) that

$\displaystyle \lim_{n\to\infty}(ax_n+by_n)=ax+by$

is true, provided $\displaystyle x_n\to x, y_n\to y$ for $\displaystyle n\to \infty$.

Now, if $\displaystyle (x_n), (y_n)$ are null-sequences, then what you know is that any linear combination $\displaystyle z_n := ax_n+by_n$ is, again a null sequence, because in that case you have $\displaystyle z_n\to z=a\cdot 0+b\cdot 0=0$.

If you know that from $\displaystyle x_n\to x$ and $\displaystyle y_n\to y$ it follows that $\displaystyle x_n y_n\to x y$, then it follows again, that the product of two null-sequences (which jus means that x=0 and y=0), is again a null-sequence $\displaystyle z_n := x_n\cdot y_n\to 0\cdot 0=0$.

Quote:

And for the product of two null-sequences is null-sequence and how we can relax conditions for the product of two null-sequences?
As to relaxing the restrictions on the product of null-sequences: if you multiply a null-sequence $\displaystyle x_n\to 0$ with a bounded sequence $\displaystyle y_n$ (which means that there exists a constant M such that for all n $\displaystyle |y_n|\leq M$), you get that the product $\displaystyle z_n := x_n \cdot y_n$ is also a null- sequence.
• Jul 20th 2010, 12:11 PM
yeKciM
Thanks (Bow)(Bow)(Bow)(Bow)(Bow) very very much !!!

(Rofl) It seems to me that's not enough... that's why i didn't post any my attempts of proofing thath one (Rofl)

It just seems to simple (Itwasntme) i don't like obviously proofs (Rofl)

THANKS!!!!