1. ## null sequence

Is sequence $a_n = (325*n^2)/(n^3 -2)$ or $b_n = (n^2 +12)/(4000n^2 +144n)$ null? Please explain.

I'm sorry if i've upset anyone by not using latex... it was my first time using this forum...and i had no idea of the house rules... please forgive.

Also, i hope you all can still help me

Thanks

2. Originally Posted by tsal15
Is sequence $a_n = (325*n^2)/(n^3 -2)$ or $b_n = (n^2 +12)/(4000n^2 +144n)$ null? Please explain.

I'm sorry if i've upset anyone by not using latex... it was my first time using this forum...and i had no idea of the house rules... please forgive.

Also, i hope you all can still help me

Thanks
"null" means the sequence converges to zero. can you tell which of these sequences, if any, converges to zero? each sequence is a ratio of polynomials. what do you know about sequences of that form?

3. if we look at them as ratio polynomials, then neither will converge....I think they will diverge from zero... am I correct?

Thanks

4. Originally Posted by tsal15
Is sequence $a_n = (325*n^2)/(n^3 -2)$ or $b_n = (n^2 +12)/(4000n^2 +144n)$ null? Please explain.
In the first one you can write,
$a_n = \frac{\frac{325}{n}}{1 - \frac{2}{n^3}}$ by dividing numerator and denominator by $n^3$.

Now the limit of this $\frac{0}{1-0} = 0$

Similarly with $b_n$

5. Originally Posted by ThePerfectHacker
In the first one you can write,
$a_n = \frac{\frac{325}{n}}{1 - \frac{2}{n^3}}$ by dividing numerator and denominator by $n^3$.

Now the limit of this $\frac{0}{1-0} = 0$

Similarly with $b_n$
I applied a similar method to $b_n = \frac{n^2 +12}{4000n^2 +144n}$
Here is my working tell me if i went wrong some where please.

$b_n = \frac{\frac{n^2 +12}{n^2}}{\frac{4000n^2}{n^2} + \frac{144n}{n^2}}$

therefore, $b_n = \frac{1 + \frac{12}{n^2}}{4000 + \frac{144}{n}}$

and when $n \rightarrow \infty$

$b_n = \frac{1 + 0}{4000 + 0}$

Leaving,
$b_n = \frac{1}{4000}$

which is 0.00025 or $2.5 * 10^{-4}$

THANKS HEAPS!!!

6. Yes.

7. so does that mean $a_n$ was the null sequence or do we assume $b_n$ is almost zero therefore we will considerate zero and thus both $a_n$ and $b_n$are null sequences?

Thanks so much Mr. PerfectHacker

8. Originally Posted by tsal15
so does that mean $a_n$ was the null sequence or do we assume $b_n$ is almost zero therefore we will considerate zero and thus both $a_n$ and $b_n$are null sequences?

Thanks so much Mr. PerfectHacker
the limit has to be zero for it to be a null sequence. so $b_n$ isn't a null sequence

Originally Posted by tsal15
if we look at them as ratio polynomials, then neither will converge....I think they will diverge from zero... am I correct?

Thanks
going back to what i said about the ratio of polynomials.

here are some rules you should get familiar with:

lets say you are given a ratio of polynomials and you want to find the infinite limit.

if the highest degree of the numerator is the same of the denominator, then the limit as $x \to \pm \infty$ to the ratio of the coefficients. (i hope you noted that when you got 1/4000)

the the degree in the denominator is greater than that of the numerator, then the ratio goes to zero as $x \to \pm \infty$

if the degree of the numerator is greater than that of the denominator, then the limit as $x \to \pm \infty$ goes to $\pm \infty$, depending on the sign of the polynomials as x gets large

9. Originally Posted by Jhevon
if the highest degree of the numerator is the same of the denominator, then the limit as $x \to \pm \infty$ to the ratio of the coefficients. (i hope you noted that when you got 1/4000)
This doesn't seem to make sense to me...

Are you trying to say that because the degrees (or powers) are the same $\frac{1^1}{4000^1}$ the limit equals the ratio/fraction as ? If I've understood it wrong, correct me please?

Everything else makes sense. thanks heaps Jhevon

10. Originally Posted by tsal15
This doesn't seem to make sense to me...

Are you trying to say that because the degrees (or powers) are the same $\frac{1^1}{4000^1}$ the limit equals the ratio/fraction as ? If I've understood it wrong, correct me please?

Everything else makes sense. thanks heaps Jhevon
the degrees of the original function, of course.

i am talking about this: $\frac {n^{\color{red}2} + 12}{4000n^{\color{red}2} + 144n}$

the highest power in the numerator and denominator is 2 for both. so the infinite limits, as $n \to \pm \infty$, goes to the ratio of the coefficients of the highest power. namely 1 in the numerator, and 4000 in the denominator. hence you get 1/4000 as your limit

11. Originally Posted by Jhevon
the degrees of the original function, of course.

i am talking about this: $\frac {n^{\color{red}2} + 12}{4000n^{\color{red}2} + 144n}$

the highest power in the numerator and denominator is 2 for both. so the infinite limits, as $n \to \pm \infty$, goes to the ratio of the coefficients of the highest power. namely 1 in the numerator, and 4000 in the denominator. hence you get 1/4000 as your limit
YES YES YES I GET IT NOW!!!!!!!!!!

THANK YOU, THANK YOU, THANK YOU, THANK YOU, THANK YOU