Question

determine if the integral 1 / (1+x^2)^1/2 converges or diverges (Hint. use comparison test)
so far i compared it to integral 1/ (x^2)^1/2 but that diverges

Answer 1

cantor u there

Answer 2

It diverges. I used the p series test though. If you distribute the exponent of 1/2, you get x^1/2, and 1/2 < 1 so it is divergent.

If you are required to use the ratio test, you must use a function that gives values smaller than the original.

Answer 3

I'm sorry not x^1/2 but it equals 1/x which if you use the integral test goes to the natural log, which is divergent. Also according to the p series test, the degree in the denominator must not equal one, so either way it diverges.

Answer 4

If you need to use the comparison test, you'd compare it to 1/x because in the example you used, you must distribute the exponent.

Answer 5

but its the wrong comparison

Answer 6

how do you compare it

Answer 7

i am using the comparison test, so 0< integral g(x) < integral f(x)

Answer 8

if integral g(x) diverges, then integral f(x) diverges , find integral g(x)

Answer 9

Yes, the function you choose must be smaller. In this instance, the +1 doesn't matter, giving us 1/(x^2)^1/2, so if you distribute the one half, you get 1/x, which we can is the function g(x).

We know g(x) diverges because of the several tests I mentioned above (both the integral, ratio, and p series tests) which means that the function f(x) which is 1/(1+x^2)^1/2, riveted as well.

Therefore, because g(x) diverges, so does f(x).

Does that answer your question?

Answer 10

no because 1/x^2 ^1/2 is greater than 1 / ( 1 + x^2 ) ^1/2

Answer 11

its in the wrong order

Answer 12

In this instance it doesn't matter, because if you take the limit as you approach infinity, the +1 doesn't matter. It's like 1/10000000 is hardly different from 1/10000001.

Answer 13

but youre not using the comparison test then

Answer 14

but i see your point

Answer 15

oh, hmmmm, dunno

Answer 16

You can check it by the Alternating series test. If it still diverges, it absolutely diverges.

Answer 17

its not an alternating series though

Answer 18

If there was an x! Or another variable, then it would change things, but a constant is irrelevant in this instance.

Answer 19

yo cantor

Answer 20

I know it's not alternating. You use the alternating test to check for absolute divergence or convergence. For example, 1/x is conditionally divergent because if it alternates, it converges. If there is no alternating piece, it diverges.

Hence, you can use the alternating test to check if it always diverges or only sometimes.

Answer 21

hmmm, how is that relevant

Answer 22

It allows you to check convergence or divergence. That was your question originally.

Answer 23

cantor

Answer 24

???????

Answer 25

yeah, this question im getting tutored for
im sorry

Answer 26

i promise i will answer your question

Answer 27

I'm saying, if you have a choice, I wouldn't use the comparison test.

If you don't, then just do the same thing with the denominator that is larger and use one of the methods I discussed earlier to see.

Answer 28

can yuo give me a formal method

Answer 29

You mean show you the steps?

To which one? The comparison test? The p series? The integral test?

Answer 30

any , the comparison test fails, as i explained earlier

Answer 31

the direct integral comparison test

Answer 32

cantor can u help me now, I have 2 parts left and I'm done. first question at the top

Answer 33

one sec