# L.C.T. Misgraded Cal. Problem?

1. Dec 8, 2004

### karen03grae

L.C.T. Misgraded Cal. Problem!???

HEy!

I just took a Cal. II test and used the Limit comparison test on one of my series. Here it is:

infinity
_
\
/ n^2/(n^3 +sqrt(n^(9)+1))
_
n=1

sorry if it looks hard to read. I compared it to 1/n. And it diverged. The teachers said I must compare it to n^2/n^(9/2). NOw doing it this way makes the series in question converge!

THE QUESTION IS WHY CAN'T I USE 1/N TO CHECK FOR CONVERGENCE USING THE LCT? Thanx, Karen

2. Dec 8, 2004

### learningphysics

Remember the LCT says that if you have two series a and b, and lim(n->infinity) a(n)/b(n)=s where s>0 and finite, then either "both a and b converge" or "both a and b diverge".

What did you get for s when you did this limit and compared to 1/n? I'm guessing you got 0 or infinity (depending on which series is on top and bottom in the ratio)? This is what I get looking at your series and 1/n. The LCT does not apply when s is 0 or infinity... so we can't conclude that since 1/n diverges the given series diverges also.

The trick is to pick a series so that when you take the ratio and then the limit, you get a finite value>0.

Its been a while since I've done this stuff. Hope this helps.

3. Dec 8, 2004

### HallsofIvy

Staff Emeritus
learningphysics is completely correct.

A little more detail: It is clear that, for large n, $\frac{n^2}{n^3+\sqrt{n^9+1}}< \frac{1}{n}$.

Unfortunately since $\Sigma \frac{1}{x}$ diverges that tells you nothing about whether $\Sigma\frac{n^2}{n^3+\sqrt{n^9+1}}$ converges or diverges. You would need to show that your terms are larger than a series that diverges to know that it diverges or smaller than a series that converges to know that it converges.

4. Dec 8, 2004

### karen03grae

I get 1/2!!!!!! I will show my work! I still believe that they both diverge.

Here it goes: Let an = n^2/(n^3 + sqrt(n^(9)+1); Let bn= 1/n.

Now: (step 1) | n^2 * n |

lim n->infinity | n^3 + sqrt(n^(9) +1) * 1 |

(step 2) | n^3*(1) |

| n^3 + sqrt(n^9(1+1/n^9))|

lim n->infinity

(step 3) |n^3*(1)|

| n^3(1+sqrt(1+1/n^9))|

lim n-> infinity

(step 4)

| 1|
___ = 1/2!!!!!!!!!!!!! AHHHHH!!!
| 1+ sqrt(1)|

<ps. sorry the divide signs would not show>

Last edited: Dec 8, 2004
5. Dec 9, 2004

### karen03grae

Can anyone explain this contradiction? I would like to believe that their are no contradicitons in the realm of mathematics. Yet, I have not seem one statement that requires one to pick the test function be the highest powers found from the series in question. With that said, I wanted my test function to be 1/n.

Any ideas?

6. Dec 10, 2004

### eddo

Simple

In step 3 I beleive it was, you took sqrt(n^9) to be n^3. This is wrong. You don't take the squareroot of the 9, you halve it, so the limit comparing to 1/n is 0, and the limit comparison test does not work for 1/n.

Since the greatest power in the numerator is 2, and the greatest in the denominator is 9/2, you must compare to 1/n^(5.2), as your teacher said, in which case you will find it to converge.

7. Dec 11, 2004

### karen03grae

Thank you Thank you Thank you!!!! You are the best!