# Thread: Does the fraction simplify?

1. ## Does the fraction simplify?

Hi all,

I've been doing well all year, but now I've hit a wall. All the exercises prior to this question have had just one term of n, but now this has shown it's ugly self, and I am absolutely confused.

The question regards sequences;

For the sequence below, decide whether it converges and, if it does, state its limit.

a(term n) = 6n^4 + 2n^3 (which is all over) 7n - 11n^4 (n=1,2,3,...)

if the fraction simplifies so I can get either the numerator or the denominator a constant with no terms of n in, I should be OK then.

Any help would be great.

Shayne.

2. Originally Posted by Hydralisk
Hi all,

I've been doing well all year, but now I've hit a wall. All the exercises prior to this question have had just one term of n, but now this has shown it's ugly self, and I am absolutely confused.

The question regards sequences;

For the sequence below, decide whether it converges and, if it does, state its limit.

a(term n) = 6n^4 + 2n^3 (which is all over) 7n - 11n^4 (n=1,2,3,...)

if the fraction simplifies so I can get either the numerator or the denominator a constant with no terms of n in, I should be OK then.

Any help would be great.

Shayne.
For fractions like this, the technique is to divide top and bottom by the highest power of n in sight (in this case, n^4), and use the fact that negative powers of n go to zero as $n\to\infty$.

3. That's helped, but I've still got terms of n in both the numerator and the denominator, so it's still "hard" for me.

I've now got:

(2/n)+ 6 (all over) (7/n^3)-11

is there anything I can do to get rid of the (2/n) in the numerator maybe?

Many thanks.

4. Originally Posted by Hydralisk
That's helped, but I've still got terms of n in both the numerator and the denominator, so it's still "hard" for me.

I've now got:

(2/n)+ 6 (all over) (7/n^3)-11

is there anything I can do to get rid of the (2/n) in the numerator maybe?

Many thanks.
You just need to see that 2/n and 7/n^3 both get smaller and smaller as n increases. In the limit, they both tend to 0 (so you can ignore them altogether), and the fraction becomes 6/(–11) (which is the answer to the question).

5. Hi,

many thanks to you both, I did infact write down 6/-11 after a bit of time thinking it through, but having it confirmed is such a releif.

Thanks again.

6. Another way to think it through is.....

As "n" approaches infinity, $6n^4$ will dwarf $2n^3$

and $-11n^4$ will dwarf $7n$

$\displaystyle-\frac{6n^4}{11n^4}=-\left(\frac{n^4}{n^4}\right)\frac{6}{11}=-\frac{6}{11}$