# null sequence

• October 25th 2008, 06:47 PM
tsal15
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
• October 25th 2008, 07:36 PM
Jhevon
Quote:

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?
• October 25th 2008, 07:51 PM
tsal15
if we look at them as ratio polynomials, then neither will converge....I think they will diverge from zero... am I correct?

Thanks
• October 25th 2008, 07:55 PM
ThePerfectHacker
Quote:

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 $b_n$
• October 27th 2008, 06:49 PM
tsal15
Quote:

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!!!
• October 27th 2008, 07:05 PM
ThePerfectHacker
Yes. (Clapping)
• October 27th 2008, 08:57 PM
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 (Wink)
• October 28th 2008, 07:04 AM
Jhevon
Quote:

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 (Wink)

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

Quote:

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
• October 28th 2008, 07:26 PM
tsal15
Quote:

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 http://www.mathhelpforum.com/math-he...cc035c3f-1.gif? If I've understood it wrong, correct me please? (Happy)

Everything else makes sense. thanks heaps Jhevon
• October 29th 2008, 04:20 PM
Jhevon
Quote:

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 http://www.mathhelpforum.com/math-he...cc035c3f-1.gif? If I've understood it wrong, correct me please? (Happy)

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
• October 29th 2008, 06:21 PM
tsal15
Quote:

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!!!!!!!!!!

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