Analysis Divergent Sequences

In summary: Also it is worth noting that there are many ways to approach this problem. In my other post I mentioned another approach that would also work.
  • #1

Homework Statement



Prove that the given sequence diverges to infinity.
{an} = (-n^4+n^3+n)/(2n+7)

Homework Equations



Diverges definition

The Attempt at a Solution



So far I have:

Let M>0 and let N= something.

I'm having a hard time figuring out what N should equal for the proof. I know I have to make the numerator smaller and denominator larger (I think), but I get confused. Can someone explain it please?
 
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  • #2
MathSquareRoo said:

Homework Statement



Prove that the given sequence diverges to infinity.
{an} = (-n^4+n^3+n)/(2n+7)

Homework Equations



Diverges definition

The Attempt at a Solution



So far I have:

Let M>0 and let N= something.

I'm having a hard time figuring out what N should equal for the proof. I know I have to make the numerator smaller and denominator larger (I think), but I get confused. Can someone explain it please?

The first thing you must notice is with that minus sign in the numerator on the ##n^4## term, this isn't going to go to ##+\infty##. It will go to ##-\infty##. Let's call your fraction ##f(n)## for the moment. Start by writing the careful definition of the statement$$
\lim_{n\rightarrow \infty}f(n) = -\infty$$Then you will know what you must prove.

[Edit, added later]: Alternatively you could show ##|f(n)|\rightarrow \infty## and observe that ##f(n)<0## for large ##n##.
 
Last edited:
  • #3
Yes, I understand that it will diverge to negative infinity. I still need help understanding how to get the N for the proof. Can you help?
 
  • #4
MathSquareRoo said:
Yes, I understand that it will diverge to negative infinity. I still need help understanding how to get the N for the proof. Can you help?

I don't know if you saw may later edit, which is the method I would use. Given ##M>0##, to find ##N## you generally use an exploratory argument which is kind of working backwards with estimates. So start with$$
\left | \frac{-n^4+n^3+n}{2n+7}\right |\ge ...$$and underestimate it to get something workable. I would start with the old calculus trick of dividing numerator and denominator by the highest power of ##n##. Then work on making the denominator larger and the numerator smaller to underestimate it with something that still goes to infinity.

[Edit]: Edited it twice. I had it right the first time. :frown:
 
Last edited:
  • #5
For sufficiently large n, the highest degree term will dominate. For sufficiently large n, (-n^4+n^3+n)/(2n+7) will be approximately -(n^3)/2.
 
  • #6
I'm sorry, I still don't understand how to write the proof for this divergent sequence. I have a similar problem done correctly I believe. Can someone help me write a proof like the following one, but for this sequence above??

Here is a similar lim, I think I have correctly proved.

{an}=(n^2+1)/(n-2)

Proof: Let M>0 and let N=2M. Then n>N implies 1/2n>M which implies (n^2+1)/(n-2)<n^2/2n=1/2n>M. Hence, lim n--> {an}=+∞.
 
  • #7
LCKurtz said:
I don't know if you saw may later edit, which is the method I would use. Given ##M>0##, to find ##N## you generally use an exploratory argument which is kind of working backwards with estimates. So start with$$
\left | \frac{-n^4+n^3+n}{2n+7}\right |\ge ...$$and underestimate it to get something workable. I would start with the old calculus trick of dividing numerator and denominator by the highest power of ##n##. Then work on making the denominator larger and the numerator smaller to underestimate it with something that still goes to infinity.

[Edit]: Edited it twice. I had it right the first time. :frown:

MathSquareRoo said:
I'm sorry, I still don't understand how to write the proof for this divergent sequence. I have a similar problem done correctly I believe. Can someone help me write a proof like the following one, but for this sequence above??

Here is a similar lim, I think I have correctly proved.

{an}=(n^2+1)/(n-2)

Proof: Let M>0 and let N=2M. Then n>N implies 1/2n>M which implies (n^2+1)/(n-2)<n^2/2n=1/2n>M. Hence, lim n--> {an}=+∞.

Did you try what I suggested? If so, how far have you gotten? Show us.
 
  • #8
I get (-n^3/2). Then I"m not sure how to solve for the N that I need.
 
  • #9
MathSquareRoo said:
I get (-n^3/2). Then I"m not sure how to solve for the N that I need.

I have no idea where you got that but, as an example of solving for ##N##, you are given an ##M## and you want ##n^{3/2} > M## for ##n > N##. So solve the inequality ##n^{3/2} > M## for ##n## to see how large ##n## needs to be. Then choose ##N## so that if ##n>N## you can guarantee ##n## is as large as it needs to be to make it work.
 
  • #10
I'm sorry, I meant (-n^3)/2. So do I set (-n^3)/2 > M and solve for n?
 
  • #11
MathSquareRoo said:
I'm sorry, I meant (-n^3)/2. So do I set (-n^3)/2 > M and solve for n?

Yes, same idea. You want the absolute value though: ##\frac{n^3}{2}>M##. So how big does ##n## need to be? What ##N## will work?
 
  • #12
N>(2M)^1/3

Is that correct?
 
  • #13
MathSquareRoo said:
N>(2M)^1/3

Is that correct?

That is how you determine how large N needs to be with an exploratory argument. Now, at this point, we have, if ##n>N## $$
\left | \frac{-n^4+n^3+n}{2n+7}\right |\ge ...\hbox{ your steps here }...\ge n^{\frac 3 2}> M$$

To write up the argument you would phrase it like this: Given ##M>0## if ##n>(2M)^\frac 1 3## then$$
\left | \frac{-n^4+n^3+n}{2n+7}\right |\ge ...\hbox{ your steps here }...\ge n^{\frac 3 2}> M$$Notice in this last paragraph, it looks like you magically came up with ##n>(2M)^\frac 1 3##, but it came from the exploratory argument.

I hope this has helped you understand the argument. Also note that since you never posted how you got the ##\frac {n^3} 2##, I am not vouching for your complete argument. I will leave that to you.
 

1. What is a divergent sequence?

A divergent sequence is a sequence of numbers that does not have a finite limit. This means that as you continue to add terms to the sequence, the numbers will continue to get larger or smaller without approaching a specific value.

2. How do you determine if a sequence is divergent?

To determine if a sequence is divergent, you can look at its limit. If the limit does not exist or is equal to infinity, then the sequence is divergent. You can also look at the terms of the sequence and see if they are increasing or decreasing without approaching a specific value.

3. What are some examples of divergent sequences?

One example of a divergent sequence is the sequence of natural numbers (1, 2, 3, 4, ...). Another example is the sequence of reciprocals of natural numbers (1, 1/2, 1/3, 1/4, ...). Both of these sequences do not have a finite limit and continue to get larger with each term.

4. Can a divergent sequence have a limit?

No, a divergent sequence cannot have a limit. A limit is a specific value that a sequence approaches as you add more terms. Since a divergent sequence does not approach a specific value, it cannot have a limit.

5. What is the importance of analyzing divergent sequences?

Analyzing divergent sequences is important in many areas of mathematics, such as calculus and number theory. It allows us to better understand the behavior of infinite sequences and can help us make predictions about their patterns and properties. Additionally, understanding divergent sequences is crucial in identifying and solving problems in real-world situations that involve infinite growth or decay.

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