Comparing Series Using the Comparison Test

[grossgermany]

Homework Statement

Does the series n^2/(n^3+n^2) diverge?

Homework Equations

We know that 1/n diverges

The Attempt at a Solution

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

Therefore intuitively it should diverge like 1/n

However, I am not very good at the Big O Small O notation. Can someone show me
1. A proof without using Big O or Small O , but purely epsilon delta, for all n>N type of argument
2. A proof involving Big O or Small O notation

Thanks

 

[╔(σ_σ)╝]

You should be trying to show that n^2/(n^3+n^2) is greater that some divergent sequence. Showing that the limit goes to zero won't help you.
But you know that,
n^3+n^2 < 2n^3. :-)

 

[Mentallic]

You can also start by simplifying it to 1/(n+1)

 

[grossgermany]

Thanks for the reply. Here is a harder question:
(n^4-10n^3)/(n^5+100n^4)

It is not obvious that we can explicitly find an N such that for all n>N the above expression is larger than 1/n

My point is that is there any rigorous way to show that if lim an= lim bn
then series an diverges if and only if series bn diverges

 

[Mentallic]

Since it's not obvious that
n^3+100n^2< 2n^3

It's not? For sufficiently large n, the higher degree polynomials are always the largest.

edit:

My point is that is there any rigorous way to show that if lim an= lim bn
then series an diverges if and only if series bn diverges

You're showing that a sequence b_n diverges while lim a_n > lim b_n so of course a_n diverges as well.

 

[╔(σ_σ)╝]

Thanks for the reply. Here is a harder question:
(n^2-100n)/(n^3+100n^2)

It is not obvious that it is greater than 1/2n^3
Since it's not obvious that
n^3+100n^2< 2n^3

It's not? For sufficiently large n, the higher degree polynomials are always the largest.

Metallic is right.

n^3 + 100n^2 <2n^3

100n^2 < 2n^3 -n^3
100n^2 < n^3
100 <n ( since n is never zero we can do this )

All we care about is sufficiently large n.

Besides you could do the following
n^3 + 100n^2 < 101n^3

:)

My point is that is there any rigorous way to show that if lim an= lim bn
then series an diverges if and only if series bn diverges[

There is a theorem or test that says if
lim (an/bn) is finite and >0 and bn diverges then an diverges.

You're showing that a sequence b_n diverges while lim a_n > lim b_n so of course a_n diverges as well.

 

[grossgermany]

Sorry I changed my question to
(n^4-10n^3+6)/(n^5+100n^4)

There is no easy quintic formula so I made up this example to deter any easy way of explicity finding the N

 

[Mentallic]

It doesn't make a difference, if you can show that (n^4-10n^3+6)/(2n^5) diverges then you can hence show that (n^4-10n^3+6)/(n^5+100n^4) diverges.

 

[╔(σ_σ)╝]

What exactly is your question? You seem to be resisting any progress.

You seem to know a lot; perhaps you can help yourself answer your questions :-).

 

[╔(σ_σ)╝]

All we are suggesting is that you should approximate your sequence by an easier one and then use an epsilon argument as freely as you want.

 

[grossgermany]

Please check my answer:
for the comparison test, claim that there exists N such that for all n>=N,
(n^4-10n^3+6)/(n^5+100n^4+999) >(n^4-10n^3+6)/(2n^5)
Proof,need the following
n^5+100n^4+999<2n^5
n+100+999/n^4 <2n
100+999/n^4 <n

Therefore N=101?

 

[grossgermany]

Is this one more difficult? How do we use the comparison test please to show the following?

sqrt[(n^8-10n^3+6)]/{sqrt[(n^7+100n^4+1)]*sqrt[n^3-500n^2+1]}