Math Question (Calc II)

[Vinnie]

Determine whether the given series converges or diverges. Sow you work and substantiate any claims:

I have three examples....

The first one is:

(infinity)
E (sigma) (n^2 +1)/(n^3 +1)
N = 1

I tried using the limit comparison test:

An / Bn and I made Bn =

(n^2 + 1)/ (n^3 +3n)

I chose this because it was similar and when you invert and multiply you end up with:

((n^2 +1)/(n^3 +1))

divided by

((n^2 + 1)/ (n^3 +3n))

= (n^3 + 3n) over (n^3 +1)

Am I on the right track?

I tried divinding all terms by highest exponent to test the limit as n--> infinity...

Not sure if I am doing what I should be....This whole series process was rushed to meet deadlines at the end of class and is confusing to me...

Any assistance would be appreciated. I'll take a detailed answer ;) but its not what I am asking for. A push in the right direction is what I would like....

Vinnie

[Barbarian]

(n^2+1)/(n^3+1) > n^2/(n^3+1) > n^2/(n^3+n^3) = 1/(2n), which diverges.

[TNorthover]

The ratio test you tried doesn't work in this case. You should be able to see that your ratio tends to 1 as n -> infinity, but the ratio test only gives you useful information if the limit is <1 or >1. Could do either for exactly 1.

Barbarian has the most obvious method for this type. He's done it properly, but the basic idea is that a quadratic divided by a cubic is roughly like 1/n for large n, which you (should) know diverges.

[Loren Pechtel]

You made me realize how much rust there is on my calc knowledge. I've got two semesters past calc II but I'm completely stumped here.

[Vinnie]

Thanks for the help. I was thinking of treating this as a limit (not sure why but I think it is legal).

As n --> infinity we have infinity over infinity.

So I can use LHopitals rule and I end up with infinity/infinity again and using Lhopital again I get 1/3n which goes off to 0.

This is the basic harmnonic series I believe and it can be proven to diverge.

Barbarian came up with 1/2n so its different from this which makes me wonder if I am doing something wrong.

Vinnie

[Vinnie]

You made me realize how much rust there is on my calc knowledge. I've got two semesters past calc II but I'm completely stumped here.

Don't feel bad. I have rust and I'm still in the course.... ;)

It always comes back with a little practice though..

[Vinnie]

The ratio test you tried doesn't work in this case. You should be able to see that your ratio tends to 1 as n -> infinity, but the ratio test only gives you useful information if the limit is <1 or >1. Could do either for exactly 1.

Barbarian has the most obvious method for this type. He's done it properly, but the basic idea is that a quadratic divided by a cubic is roughly like 1/n for large n, which you (should) know diverges.

I didn't think I was using a ratio test. I thought I was getting at a lim. comparison test. I don't think we got to the ratio test. We didn't have enough time for the last two methods but I did look up the ratio test and it seems that is what Barbarian did do and it makes a lot of sense.

Vinine

[Vinnie]

For something such as:

(infinity)
E (sigma) (e/Pi)^(n-1)
n= 1

Can I just treat the r as (e/pi)^(n-1)
and assume or factor out a 1 as the a?

And use the ar^(n-1) form?

I know the approximate value of r which is 2.71/3.14 = < 1
thus series converges to sum of a/(1-r)

Thus it converges to 1/(1- (2.71/3.14))

Is that the correct approach??? It just seems too easy, like I shouldn't be assuming the a = 1 up above.

I am use to another form like 7/3n where you factor it out so it equals ar^(n-1)
But this was already written in that form.

Vinnie

[premjan]

I don't think you can use L'Hopital's rule, because it only gives the value of the limit whereas you are looking for the sum. The limit could go to zero but the sigma still be infinite (as in the case of the harmonic series).

[Barbarian]

Barbarian came up with 1/2n so its different from this which makes me wonder if I am doing something wrong.Well, I don't think you are doing something wrong, you are doing something else.

Convergence or divergence of \sum_{n=0}^\infty \frac{n^2+1}{n^3+1} is in question.

We start by observing that \frac{n^2+1}{n^3+1} \; > \; \frac{n^2}{n^3+1} \; > \; \frac{n^2}{n^3+n^3} \; = \; \frac1{2 \; n}

therefore

(1) \sum_{n=0}^m \frac{n^2+1}{n^3+1} \; > \; 1 + \frac12 \sum _{n=1}^m \frac1{n} \, \forall m \in \bb{N}, \, m \; > \; 0

Since we know that \sum _{n=1}^m \frac1{n} diverges, we can state that \forall \epsilon \in {\bb{R}} \, , \, \exists k\(\epsilon\) \in \bb{N} such that for \forall m>k\(\delta\) we have \sum_{n=1}^m \frac1{n} \; > \; \epsilon. This can also be restated for replacing \epsilon with 2 \; \(\delta \; - \; 1\) where \delta is an arbitrary real: for any such \delta, we can find a k\(\delta\) \in {\bb{N}} such that \forall m>k\(\delta\) we have \sum_{n=1}^m \frac1{n} \; > \; 2 \; \(\delta \; - \; 1\).

Comparing this with (1), we get that for \forall \delta \in {\bb{R}}, \, \exists k\(\delta\) \in \bb{N} such that for \forall m > k\(\delta\) we have

\sum_{n=0}^m \frac{n^2+1}{n^3+1} \; > \; 1 + \frac12 \sum _{n=1}^m \frac1{n} \; > \; 1 \; + \; \frac12 \( 2 \; \(\delta \; - \; 1 \) \) \; = \delta

i.e. the formal statement of divergence (towards +\infty).

This is not the ratio test. I did not try to use any such indirect rules, as they all fail for the related problem \sum _{n=1}^\infty \frac1{n}.

[Sven]

Thanks for the help. I was thinking of treating this as a limit (not sure why but I think it is legal).

As n --> infinity we have infinity over infinity.

So I can use LHopitals rule and I end up with infinity/infinity again and using Lhopital again I get 1/3n which goes off to 0.

This is the basic harmnonic series I believe and it can be proven to diverge.

Barbarian came up with 1/2n so its different from this which makes me wonder if I am doing something wrong.

I spotted one problem with your approach:

L'Hopital is normally used only for continuous variables (that is, for fractions of the form lim (x->x0) f(x)/g(x), with both f(x0) = g(x0) = 0 or both f(x) and g(x) diverging).

I really have no idea if L'Hopital also works in general for discrete series. In your case, it worked. But one example does not make a rule.

[Sven]

For something such as:

(infinity)
E (sigma) (e/Pi)^(n-1)
n= 1

Can I just treat the r as (e/pi)^(n-1)
and assume or factor out a 1 as the a?

And use the ar^(n-1) form?

I know the approximate value of r which is 2.71/3.14 = < 1
thus series converges to sum of a/(1-r)

Thus it converges to 1/(1- (2.71/3.14))

Is that the correct approach??? It just seems too easy, like I shouldn't be assuming the a = 1 up above.

Looks about right to me. It is that easy. :)

[premjan]

LHopital won't work because it just gives you the limit of the expression which goes to zero, but it doesn't tell you whether the series as a whole is converging or diverging - in this case it diverges, but I don't think Lhopital tells you that.
http://en.wikipedia.org/wiki/L'Hôpital's_rule

[Preno]

I spotted one problem with your approach:

L'Hopital is normally used only for continuous variables (that is, for fractions of the form lim (x->x0) f(x)/g(x), with both f(x0) = g(x0) = 0 or both f(x) and g(x) diverging).

I really have no idea if L'Hopital also works in general for discrete series. In your case, it worked. But one example does not make a rule.If the limit x->inf f(x) exists, then the limit n->inf f(n) must exist and is equal to that of f(x), by the definition of limit.

[Vinnie]

I saw what Barbarian did in my book. It was in section 10.4, I don't think we got there but I ended up just doing that one that way and did a term by term comparison and put in the official notation. They called it the Direct Comparison test and did it for a few examples.

Anyways, my approach to the last one (I had three take home problems counting towards the final, 6 points each out of a doublt test = 200 points).

My last one was:

(infinity)
E (sigma) (n^2)(e^(-n^3))
n=1

I'm hoping my basic algebra rewrite was//is correct/

I rewrote it as

n^2/ (e^(n^3))

Thus getting rid of the negative exponent.

Next I used a u subsition of u = n^3 which got rid of the denominator: =

1/3 Integral of du / e^u

I brought the e back to the numerator =

1/3 the integral of e^-u du

and integrated it as

-e^-u / 3.

From here I did what the book did and treated it as a limit

lim
n-->b of (-e^-u) / 3 over interval of 1 to b
edited to add: (have to replace u = n^3)


If I plugged in f(b) - f(a) correctly I ended up with 0 t 1/e and that Converges.



Did I do this correctly? I hope so, I have to hand it in tomorrow when I take the final. I'm just really happy this portion of the final was a takehome because I would have completely bombed it on the final!

Vinnie

[Vinnie]

Looks about right to me. It is that easy. :)

Thanks. Now I just hope my teacher agrees with us :D

I should be on a picnic with my girlfriend but I have to go gind out some more review problems.....:rolleyes: :huh:

Vinnie

[Sven]

If the limit x->inf f(x) exists, then the limit n->inf f(n) must exist and is equal to that of f(x), by the definition of limit.
Could you expand on this?

Even if true (I suspect it is, you seem to know more on this), does this mean automatically mean that L'Hopital works here? (which was my point...!)

[Sven]

I saw what Barbarian did in my book. It was in section 10.4, I don't think we got there but I ended up just doing that one that way and did a term by term comparison and put in the official notation. They called it the Direct Comparison test and did it for a few examples.

Anyways, my approach to the last one (I had three take home problems counting towards the final, 6 points each out of a doublt test = 200 points).

My last one was:

(infinity)
E (sigma) (n^2)(e^(-n^3))
n=1

I'm hoping my basic algebra rewrite was//is correct/

I rewrote it as

n^2/ (e^(n^3))

Thus getting rid of the negative exponent.

Next I used a u subsition of u = n^3 which got rid of the denominator: =

1/3 Integral of du / e^u
Umm, how did you manage to replace the sum by an integral?

[premjan]

If the integral converges, I think the sum will converge as well.

[premjan]

Should that be 1/3e instead of 1/e at the end?

[Preno]

Could you expand on this?

Even if true (I suspect it is, you seem to know more on this), does this mean automatically mean that L'Hopital works here? (which was my point...!)Well, to say that lim x->inf f(x) = a is to say that for every e>0 there exists a K such that for all x>K, |f(x)-a| < e. To say that the limit of a sequence a_i is a is to say that for every e>0 there exists a K such that for all n>K, |a_n - a| < e. If something holds for all real numbers >K, it also holds for all naturals >K. Thus, if the limit of f(x) as a function exists, then the limit of f(n) as a sequence exists and is equal to it.

L'Hopital may or may not work for determining the limit of a function, but if it is valid to use it for the function, then the limit of the sequence thus can be determined. Thus, whether one can use l'Hopital for inf/inf or not does not depend on whether the original was a sequence or a function. Of course, one cannot just apply l'Hopital to a series directly. But if you determine that lim (a_n)/(1/n) = c (c != 0) using l'Hopital, then the series Sum(a_n) should diverge because Sum(1/n) diverges.

[so-crates]

Is there any theory which says that the series sum(f(n), n = 0 -> inf would converge if and only if lim( f(n) - f(n-1)) -> 0 as n-> inf? That seems right, and I can't think of a counter example off the top of my head.

[Preno]

No, that's incorrect. 1/n is a simple counterexample. The latter is a necessary but not sufficient condition.

[EthnAlln]

Determine whether the given series converges or diverges. Sow you work and substantiate any claims:

I have three examples....

The first one is:

(infinity)
E (sigma) (n^2 +1)/(n^3 +1)
N = 1

I tried using the limit comparison test:

An / Bn and I made Bn =

(n^2 + 1)/ (n^3 +3n)

I chose this because it was similar and when you invert and multiply you end up with:

((n^2 +1)/(n^3 +1))

divided by

((n^2 + 1)/ (n^3 +3n))

= (n^3 + 3n) over (n^3 +1)

Am I on the right track?

I tried divinding all terms by highest exponent to test the limit as n--> infinity...

Not sure if I am doing what I should be....This whole series process was rushed to meet deadlines at the end of class and is confusing to me...

Any assistance would be appreciated. I'll take a detailed answer ;) but its not what I am asking for. A push in the right direction is what I would like....

Vinnie


Limit comparison with Sum 1/n.

[Sven]

Well, to say that lim x->inf f(x) = a is to say that for every e>0 there exists a K such that for all x>K, |f(x)-a| < e. To say that the limit of a sequence a_i is a is to say that for every e>0 there exists a K such that for all n>K, |a_n - a| < e. If something holds for all real numbers >K, it also holds for all naturals >K. Thus, if the limit of f(x) as a function exists, then the limit of f(n) as a sequence exists and is equal to it.
Hmm, OK, if we view the sum as a series, this should work. Thanks!

L'Hopital may or may not work for determining the limit of a function, but if it is valid to use it for the function, then the limit of the sequence thus can be determined. Thus, whether one can use l'Hopital for inf/inf or not does not depend on whether the original was a sequence or a function. Of course, one cannot just apply l'Hopital to a series directly. But if you determine that lim (a_n)/(1/n) = c (c != 0) using l'Hopital, then the series Sum(a_n) should diverge because Sum(1/n) diverges.
My problem is that L'Hopital uses derivatives - which are not defined for sequences, but only for functions (to my knowledge).

[so-crates]

No, that's incorrect. 1/n is a simple counterexample. The latter is a necessary but not sufficient condition.


I dont' see how 1/n would be a counter example since 1/n - 1/n-1 = n-1/n(n-1) - n/n(n-1) = -1/n(n-1) which limit is 0 as n->inf

[Preno]

Yes. And sum (1/n) diverges.
My problem is that L'Hopital uses derivatives - which are not defined for sequences, but only for functions (to my knowledge).First you broaden the sequence to a function including non-natural arguments, then you determine its limit, and thenn since if that function has a limit, then so does the sequence and they must be the same, you can say that this is also the limit of the sequence.

[Sven]

First you broaden the sequence to a function including non-natural arguments, then you determine its limit, and thenn since if that function has a limit, then so does the sequence and they must be the same, you can say that this is also the limit of the sequence.
Perhaps my problem is that I've forgotten why L'Hopital actually works. :redface:

[premjan]

L'Hopital is provable by the mean value theorem. Where the function is 0 at the endpoint of an interval, the derivative is proportional to the function.