Comparing Series Using the Comparison Test

  • Thread starter grossgermany
  • Start date
In summary: There is no easy quintic formula so I made up this example to deter any easy way of explicity finding the NIt 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.What exactly is your question? You seem to be resisting any progress.
  • #1
grossgermany
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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
 
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  • #2
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. :-)
 
  • #3
You can also start by simplifying it to 1/(n+1)
 
  • #4
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
 
Last edited:
  • #5
grossgermany said:
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:
grossgermany said:
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 bn diverges while lim an > lim bn so of course an diverges as well.
 
  • #6
grossgermany said:
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

Mentallic said:
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 bn diverges while lim an > lim bn so of course an diverges as well.
 
  • #7
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
 
  • #8
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.
 
  • #9
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 :-).
 
  • #10
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.
 
  • #11
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?
 
Last edited:
  • #12
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]}
 
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1. What does it mean for a series to diverge?

When a series diverges, it means that the sum of its terms increases without bound as you add more and more terms. In other words, the series does not have a finite sum and will continue to grow infinitely.

2. How can I determine if a series diverges?

One way to determine if a series diverges is to use a convergence test. Some commonly used tests include the comparison test, ratio test, and integral test. These tests can help determine if a series converges or diverges.

3. What is n^2/(n^3+n^2)?

n^2/(n^3+n^2) is a mathematical expression that represents a series. The variable n represents the number of terms in the series, and n^2 represents the sum of the squares of those terms while n^3 represents the sum of the cubes of those terms.

4. How can I prove that n^2/(n^3+n^2) diverges?

To prove that a series diverges, you can use a convergence test or mathematical proof. In the case of n^2/(n^3+n^2), we can use the limit comparison test to show that the series diverges as it is comparable to the harmonic series, which is known to diverge.

5. Are there any real-world applications of n^2/(n^3+n^2)?

n^2/(n^3+n^2) may have applications in various fields, such as physics and engineering, where series and convergence tests are used to analyze data and make predictions. However, it is essential to carefully consider the context in which the series is being used and to understand its limitations and assumptions.

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