# Math Help - Converging Sequence

1. ## Converging Sequence

I got a questions:

How can i tell if the following sequence is converging and if it is how do i find the limit?
a n = (9+2n^6)/(n+n^6)

Thanks

2. Hello,

Originally Posted by taurus
I got a questions:

How can i tell if the following sequence is converging and if it is how do i find the limit?
a n = (9+2n^6)/(n+n^6)

Thanks
A sequence converges if its limit when n tends to infinity exists.

Here :

\begin{aligned} \lim_{n \to \infty} a_n &=\lim_{n \to \infty} \frac{9+2n^6}{n+n^6} \ \text{divide both by } n^6 \\ \\
&=\lim_{n \to \infty} \frac{\frac{9}{n^6}+2}{\frac{1}{n^5}+1} \end{aligned}

But :
$\lim_{n \to \infty} \frac{9}{n^6}=0$
And :
$\lim_{n \to \infty} \frac{1}{n^5}=0$

Hence :

$\lim_{n \to \infty} a_n=\frac{2}{1}=\boxed{2}$

Is it what you wanted ?

3. Originally Posted by taurus
I got a questions:

How can i tell if the following sequence is converging and if it is how do i find the limit?
a n = (9+2n^6)/(n+n^6)

Thanks
We know they converge because the two polynomials in the numerator and denominator have the same degree. We find the limit by dividing each term by $n^6$.

$a_n=\frac{9+2n^6}{n+n^6}$

$\lim_{n\to\infty}\frac{9+2n^6}{n+n^6}$

$\lim_{n\to\infty}\frac{\frac{9}{n^6}+\frac{2n^6}{n ^6}}{\frac{n}{n^6}+\frac{n^6}{n^6}}$

$\lim_{n\to\infty}\frac{\frac{9}{n^6}+2}{\frac{1}{n ^5}+1}$

$\frac{0+2}{0+1}$

$a_{\infty}=2$

4. Few questions:

1) how come 9/n^6 = 0 and same for 1/n^5?
2) so n/n^6 is equal to 1/n^5?
3) why choose n^6?

thanks`

5. Originally Posted by taurus
Few questions:

1) how come 9/n^6 = 0 and same for 1/n^5?
$\frac{9}{n^6}$ is not zero; $\lim_{n\to\infty}\frac{9}{n^6}$ is zero. We know this because as $n$ approaches $\infty$, the denominator approaches $\infty$, while the numerator remains a constant. And thus, because $\frac{9}{\infty^6}=0$, the fraction $\frac{9}{n^6}$ approaches zero as $n$ approaches $\infty$.

2) so n/n^6 is equal to 1/n^5?
Basic algebra.

$\frac{n}{n^6}=\frac{n}{n*n*n*n*n*n}=\frac{1}{n*n*n *n*n}=\frac{1}{n^5}$

3) why choose n^6?
Anything greater and we get a zero in the denominator of the whole thing, which leaves the solution undefined.

6. Originally Posted by taurus
3) why choose n^6?
Because it is the highest power in the term.

Consider if we had done 1/n^5

$\lim_{n\to\infty} \frac{9+2n^6}{n+n^6}$

$=\lim_{n\to\infty} \frac{\frac 9{n^5}+2n}{\frac 1{n^4}+n}$

Well, that's better than we had before, but still not as blatantly obviously straight forward as using n^6, because in this one, the numerator and denominator still both go to infinity, which means you will have to do more simplifying.

--------------------

Consider if we used 1/(n^7)

$\lim_{n\to\infty} \frac{9+2n^6}{n+n^6}$

$=\lim_{n\to\infty} \frac{\frac 9{n^7}+\frac 2n}{\frac 1{n^6}+\frac 1{n}}$

On this one, the numerator and denominator both go to zero, so you will again have to do more fiddling with it.

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But if we wisely choose the highest degree in the term, then it comes out nicely, observe if we use 1/(n^6)

$\lim_{n\to\infty} \frac{9+2n^6}{n+n^6}$

$=\lim_{n\to\infty} \frac{\frac 9{n^6}+2}{\frac 1{n^5}+1}$

Now our numerator equals 2, and our denominator equals 1, so we don't have to do anything more with it, we can apply the limit finally ^_^

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Here are two very similar examples of how you would choose the term to multiply by:
$\lim_{n\to\infty} \frac{9+2n^4}{n+n^5}$

On this one, the greatest degree of n is 5, so multiply the numerator and denominator by 1/n^5

$=\lim_{n\to\infty} \frac{\frac 9{n^5}+\frac 2n}{\frac 1{n^4}+1}$

Look, the numerator goes to zero, and the denominator goes to 1, so we can apply our limit, the sequence converges to zero hooray!

$\lim_{n\to\infty} \frac{9+2n^8}{n+n^7}$

In this one, our highest degree of n is 8, so we multiply numerator and denominator by 1/(n^8)

$=\lim_{n\to\infty} \frac{\frac 9{n^8}+2}{\frac 1{n^7}+\frac 1n}$

Uh oh, in this one, the numerator approaches 2, but the denominator approaches zero. And we know that when we divide something by smaller and smaller numbers, it gets larger and larger (ie 16/8=2, 16/4=4, 16/2=8, 16/1=16, 16/.5=32, 16/.25=64 and so on). So as our denominator approaches zero, our numerator approaches infinity, so the sequence diverges.

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If we had chosen any other value than the highest power of n, then we would have to do more arranging and fiddling to get it into a form we could find the answer to, so that is why we chose n^6 on your sequence, that is why we chose n^5 on the first example, and that is why we chose n^8 on the second example. As you get more familiar with these, you may find that you see a situation where some other value will work equally well or better, (such as n^7 on the second example, which will also render an answer, and may be easier for you to see even) but in most cases (at least most that you're likely to see in your class), choosing the highest power of the equation will treat you well.