# Homework Help: Comparison Test

1. Jun 4, 2010

### estro

I have hard time with latex so I attached all relevant info in that picture file.
Thanks you in advance!

Last edited: Jun 4, 2010
2. Jun 4, 2010

### System

Do not make it complicated

take the limit(as n goes to infinity) of both sides of the inequality in the question (the inequality which has the nth rooth of the absolute value of an)

the left side will be simply the limit which is in the root test and the right side will be 1
so the limit of the nth root of the absolute value of an is less than 1
so the series converges by the nth root test.

3. Jun 4, 2010

### estro

I'm not sure if I get you right.
The root test tell us something only if the limit is not 1, but in our case the limit is 1.

4. Jun 4, 2010

### System

No.
Take the limit of both sides as I said, the limit of the right side will be 1
so the limit of the right side (which the nth root of the absolute value of an) is LESS THAN 1 ---> The series converges by the root test.

5. Jun 4, 2010

### estro

Sorry, but I still can't get the idea.

You mean this expression? How you conclude that it is less then 1?

6. Jun 4, 2010

### System

Sir,
The inequality itself

take the limit of both sides

it will something like
the lim of the left side LESS THAN the lim of the right side

7. Jun 4, 2010

### estro

Where in the inequality you see "LESS THAN"? (It's less OR EQUAL to)
OBVIOUSLY, the root test CAN'T be used here.

Last edited: Jun 4, 2010
8. Jun 4, 2010

### System

yeah, its less than or equal
Wait, I will do it in a picture.

9. Jun 4, 2010

### System

Its in the attachment.

#### Attached Files:

• ###### convergent.JPG
File size:
9.2 KB
Views:
121
10. Jun 4, 2010

### estro

11. Jun 4, 2010

### System

your website said: if L < 1 then the series absolutely convergent (hence convergent).
Its agree with my conclusion :)

12. Jun 4, 2010

### System

If you mean less than or EQUAL , then you could consider the inequality starts from n=2, and this will be remove the "EQUAL" sign from the inequality
Finally, you know that the finite sum does not affect the convergence of the series, so it will include the 1.

13. Jun 4, 2010

### estro

The limit is LESS OR EQUAL to 1, means the limit CAN be 1, which means you CAN'T use the root test!

What?! (we are talking about LIMIT)

14. Jun 4, 2010

### System

If f(n)<g(n) for n>a

then:

f(n)<g(n) for n>a+1

15. Jun 4, 2010

### estro

Thank you for your help, but I'll keep looking for other ideas (since I believe your idea is wrong).

16. Jun 4, 2010

### Red_CCF

I'm not sure if this is right but what I did was:

let bn = 1 - 1/n^a, for 0 < a < 1, and an = bn ^ n; by virtue of the root test, this converges since the limit is between but not equal to 0 and 1.

17. Jun 4, 2010

### estro

I'm not sure but I think it is not. Again root test can't be used here.(despite the resemblance)

18. Jun 4, 2010

### Dick

Good for you for resisting all of the bad root test advice. You could try to apply an integral test to exp(-n^(1-a)). It gives you an incomplete gamma function, if you know how to deal with that. I mostly know this because I typed it into Maxima. It diverges as a->1, of course, but seems to be finite on a in (0,1).

19. Jun 5, 2010

### estro

Thank you for your response.

I'm not too familiar with the gamma function and how to deal with it,
nevertheless I tried applying integral test to exp(-n^(1-a)), ( too much work=) and not sure how to integrate it anyway).
Do you think there is no better way, and I should make it with the integral test?

Last edited: Jun 5, 2010
20. Jun 5, 2010

### estro

Intuition suggests getting from exp(-n^(1-a)) to geometric series, but I can't figure out algebra recipe to get there... =(
Any ideas?

21. Jun 5, 2010

### Dick

The integral test isn't that bad, especially if you remember that you don't need to find an exact value for the integral. You just need to show the integral converges. Change variables to u=n^(1-a). Can you show the integral is u^K*exp(-u) for some constant K?

22. Jun 5, 2010

### estro

Thank you for your guidance.

Here is my raw attempt, I'm on right way? (My integration is a little nasty...)

If all I did is ok it should be trivial from where I stopped.

#### Attached Files:

• ###### word document.zip
File size:
10.1 KB
Views:
27
Last edited: Jun 5, 2010
23. Jun 5, 2010

### Dick

Double check that. If you do it right the new integral should be t^(a/(1-a))*exp(-t). That means the power of t is positive, so t^(a/(1-a)) never becomes less than 1. You want to show this converges by showing exp(-t) goes to zero fast enough. exp(t) grows faster than any polynomial in t as t->infinity, right?

24. Jun 5, 2010

### estro

Oh, I blundered with my integration.
But now how can I show that Integral_(t^(a/1-a)/e^t)dt from 1 to infinity is convergent?

25. Jun 5, 2010

### Dick

Call K=(a/(1-a)). For any N you can find t large enough so that exp(t)>t^N. Pick N=K+2 for example.

Last edited: Jun 5, 2010