# Homework Help: Limit comparison test

1. Sep 20, 2010

### ProPatto16

1. The problem statement, all variables and given/known data

show whether $$\sum$$ (n+5)/(n3-2n+3) is convergant of divergant

2. Relevant equations

limit comparison test, lim an/bn = c where c > 0

3. The attempt at a solution

an is given
let bn = n/n3

so then:

lim (n+5)/(n3-2n+3)
n/n3
= lim (n+5)n3 / (n3 - 2n + 3)n
= lim (n4 + 5n3) / (n4 - 2n2 +3n)
= lim (5n3) / (-2n2+3n)

now as n -> infinity, lim (5n3) / (-2n2+3n) -> -5n/2

and as n -> inifinity, -5n/2 -> -inifinity... therefore divergant? since c < 0??

feel free to tell me i have no idea. these things confuse me so much.

2. Sep 20, 2010

### Staff: Mentor

Instead of bn = n/n3, simplify this to 1/n2.

The limit you should be working with is
$$\lim_{n \to \infty} \frac{\frac{n + 5}{n^3 - 2n + 3}}{\frac{1}{n^2}}$$

3. Sep 20, 2010

### ProPatto16

that limit simplifies to

lim 5n2/(-2n+3)

so then my previous explantion still stands...

now as n -> infinity, lim (5n2) / (-2n+3) -> -5n/2

and as n -> inifinity, -5n/2 -> -inifinity... therefore divergant? since c < 0??

so i dunno...

i also tried taking the simplified limit and dividing through by n2 which gives

lim 5/(-2/n + 3/n2) and i know the limit of 1/n or 1/n2
and all varieties is 0... so then

lim 5/(-2/n + 3/n2) = 5/(0-0) = 0... since that is not >0... then it hasnt worked...

4. Sep 20, 2010

### snipez90

No, that is incorrect. Remember for a rational function f(n), if you let n go to infinity only the terms with the largest exponent on n remain in the numerator and in the denominator.

For instance, lim (5n^2 + 1000n)/(3n^2 + 40n + 500) is simply lim (5n^2)/(3n^2) = 5/3. Intuitively as n grows beyond all bounds, 5n^2 completely dominates 1000n in the numerator, and similarly 40n + 500 is insignificant compared to 3n^2 when n is approaching infinity.

5. Sep 20, 2010

### Staff: Mentor

Check your algebra. For large n, the numerator grows at exactly the same rate as the denominator.

6. Sep 20, 2010

### ProPatto16

yeah i used that method of thinking to reduce the limit down:

so then lim 5n2/(-2n+3) becomes lim 5n/-2

but then i dunno where to go from there..

7. Sep 20, 2010

### Staff: Mentor

That's still wrong. The limit is a finite constant.

8. Sep 20, 2010

### ProPatto16

mark... yeah i did.... it all comes down to a ratio of -2.5n... so the limit still dpeends on n.

if n = 1000000000 then it comes to -2500000004... i can see the effect here... i just cant summarise it in a way that gives me an appropriate answer...

9. Sep 20, 2010

### ProPatto16

the only finite constant here that would make any relevant sense is -5/2...

10. Sep 20, 2010

### Staff: Mentor

No, it does not depend on n.

No.

The numerator is (n + 5)/(n^3 - 2n + 3). The denominator is 1/n^2. For large n, the numerator is roughly n/n^3 or 1/n^2.

11. Sep 20, 2010

### ProPatto16

(n+5)/(n3-2n+3) is the numerator. for large values of n... this becomes n/n3

the denominator is 1/n2

the numerator can simplify down to 1/n2 also.

so then:

lim (n+5)/(n3-2n+3) / 1/n2

becomes lim 1/n2 / 1/n2

= 1
since this is greater than 0... both functions converge or diverge... and since 1/n2 converges... so does (n+5)/(n3-2n+3) ??

12. Sep 20, 2010

### Staff: Mentor

Which is 1/n^2.
YES!!!
Yes.

13. Sep 20, 2010

### ProPatto16

thank you. sorry for being frustrating.

14. Sep 20, 2010

### Staff: Mentor

Here are the steps in evaluating that limit.
$$\lim_{n \to \infty} \frac{\frac{n + 5}{n^3 - 2n + 3}}{\frac{1}{n^2}}$$
$$=\lim_{n \to \infty} \frac{n + 5}{n^3 - 2n + 3} \frac{n^2}{1}$$
$$=\lim_{n \to \infty} \frac{n(1 + 5/n)}{n^3(1 - 2/n^2 + 3/n^3} \frac{n^2}{1}$$
$$=\lim_{n \to \infty} \frac{(1 + 5/n)}{(1 - 2/n^2 + 3/n^3} \frac{1}{1}$$
= 1

I did quite a bit of cancelling after factoring out n, n^3, and so forth. All of the terms with n to some power in the denominator go to zero when n gets large.

15. Sep 21, 2010

### ProPatto16

i see... factorise out the highest exponent that appears in the equation.

thanks a lot:)