Just a bit of help understanding convergance

P

Probability

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I'll try and keep this as simple as I can.

We have say;

(15n - 4n) / (3n + 12) (n = 1,2,3...)

If

(15(1) - 4(1)) = 11

and

(3(1) + 12)) = 15

then

(11) / (15) = 0.73

So this is where the confusion starts for me, is it the final answer 0.73 that decides if the sequence converges or is it the answer to each section of the sequence, i.e. the answer to the numerator does or does not converge, and or the denominator does converge or does not?

Thanks
 
Sorry if i didn't understand your question,but
substituting 1 into the sequence doesn't really show anything about convergence.
Your sequence is 11n/(3n+12).0.73 is just the first term of the sequence.For deciding if it converges or not you can use many methods,easiest should be saying that if n is "large" then the sequence is 11n/3n=11/3 so it converges to 11/3.
 
Probability said:
I'll try and keep this as simple as I can.

We have say;

(15n - 4n) / (3n + 12) (n = 1,2,3...) ............ and what do you have to do with this sequence?

If

(15(1) - 4(1)) = 11

and

(3(1) + 12)) = 15

then

(11) / (15) = 0.73

So this is where the confusion starts for me, is it the final answer 0.73 that decides if the sequence converges or is it the answer to each section of the sequence, i.e. the answer to the numerator does or does not converge, and or the denominator does converge or does not?

Thanks
Click to expand...

Your post is incomplete and does not make sense.
 
OK I'll try again.

a_n = (12 - 8n) / (4n + 36) (n = 1,2,3...)

I am looking at trying to understand convergence with these types of sequences and their limits. Given that this is a brand new topic to me I am blinded with information that is OTT for a beginner.

I assume that the test for convergence means that a_n tends to a definite value as n tends to infinity.

I am assuming that the sequence is n = 1,2,3..., therefore n will increase.

If the other member who posted is correct (I have little understanding at the moment) then;

12 - 8 = 4n, and

4 + 36 = 40n, therefore

4n / 40n

In which case I assume based on the product rule and the sequence n = 1.2.3..., that this sequence will not converge and will not tend to 0.

Please advise.

Thanks
 
Probability said:
OK I'll try again.

a_n = (12 - 8n) / (4n + 36) (n = 1,2,3...)

I am looking at trying to understand convergence with these types of sequences and their limits. Given that this is a brand new topic to me I am blinded with information that is OTT for a beginner.

I assume that the test for convergence means that a_n tends to a definite value as n tends to infinity.

I am assuming that the sequence is n = 1,2,3..., therefore n will increase.

If the other member who posted is correct (I have little understanding at the moment) then;

12 - 8 = 4n, and

4 + 36 = 40n, therefore

4n / 40n

In which case I assume based on the product rule and the sequence n = 1.2.3..., that this sequence will not converge and will not tend to 0.

Please advise.

Thanks
Click to expand...

You need to calculate:

\(\displaystyle \displaystyle{\lim_{n \to \infty} \left [\dfrac{12 - 8n}{4n +36}\right ]}\)

Now if this limit is a "finite number" → the "sequence" is convergent.

You know how to calculate the above limit - correct?
 
At the moment I only know what I have typed to date, so if I am wrong on that then I don't know?
 
Can anyone point me in the right direction please with regards methods to work these sequences out and finding the limits and converges.

I have two examples;

1/ a_n = (12 - 8n) \ (4n + 36)

do I do it this way;

(12 - 8n) = 4n and (4n + 36) = 40n so that I end up with (4n) \ (40n)

Thus the n values are increasing and therefore are not converging, therefore the limit being 4/40

Thanks
 
Do you understand that \(\displaystyle \displaystyle{\lim _{n \to \infty}}\dfrac{c}{n} = 0~?\)

If you do then \(\displaystyle \displaystyle{\lim _{x \to \infty }}\dfrac{{12 - 8n}}{{4n - 36}} = {\lim _{x \to \infty }}\dfrac{{3 - 2n}}{{n - 9}} = {\lim _{x \to \infty }}\frac{{\frac{3}{n} - 2}}{{1 - \frac{9}{n}}} = - 2\)
 
pka said:
Do you understand that \(\displaystyle \displaystyle{\lim _{n \to \infty}}\dfrac{c}{n} = 0~?\)

If you do then \(\displaystyle \displaystyle{\lim _{x \to \infty }}\dfrac{{12 - 8n}}{{4n - 36}} = {\lim _{x \to \infty }}\dfrac{{3 - 2n}}{{n - 9}} = {\lim _{x \to \infty }}\frac{{\frac{3}{n} - 2}}{{1 - \frac{9}{n}}} = - 2\)
Click to expand...

Thanks for replying pka.

No I don't understand \(\displaystyle \displaystyle{\lim _{n \to \infty}}\dfrac{c}{n} = 0~?\) this, it is a very new subject to me today. I can follow your method of working thereafter but fail to see how you reached the limit - 2

I think I need somebody to walk me through the fundamentals of this branch of sequences as the books I have don't explain in in laymans terms.

Please advise if you will.
 
Probability said:
Thanks for replying pka.

No I don't understand \(\displaystyle \displaystyle{\lim _{n \to \infty}}\dfrac{c}{n} = 0~?\) this, it is a very new subject to me today. I can follow your method of working thereafter but fail to see how you reached the limit - 2

I think I need somebody to walk me through the fundamentals of this branch of sequences as the books I have don't explain in in laymans terms. Please advise if you will.
Click to expand...

Think about \(\displaystyle \displaystyle{\lim _{n \to \infty }}\frac{{12}}{n}\) you are dividing 12 by a ever growing number, n is getting bigger and bigger.
So that means \(\displaystyle \dfrac{{12}}{n}\) is getting closer and closer to 0

Thus \(\displaystyle \displaystyle{\lim _{n \to \infty }}\frac{{12}}{n}=0\)
 
I have two rules to work with.

1/ Reciprocal Rule

If the terms of a sequence b_n are of the form 1/a_n, where terms of the sequence a_n become arbitrarily large as n increases, then the sequence b_n is convergent, and the lim n tends to infinity b_n = 0

2/ Constant Multiple Rule

If the terms of a sequence b_n are of the form ca_n, where the sequence a_n is convergent with limit 0, and c is a constant, then the sequence b_n is also convergent, and lim n tends to infinity b_n = 0

so

b_n = (3^n - 7) / (15(0.4)^n + 3)

Because (3^n - 7) is becoming arbitrary large, this would not converge to 0, am I right on that point?

Because (15(0.4)^n + 3) i.e. (0.4)^n is less than 1, this will converge to 0.

So would I be right in saying that this sequence will not converge, and if this is so how do I work out the limit?

Would this is correct.

b_n = (3^n - 7) / (15(0.4)^n + 3)

(3^1 - 7) = - 4, and

(15(0.4)^1 + 3) = 9, therefore

(- 4) / (9) = - 0.44 (2dp)

Am I a mile off?
 
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Probability said:
Am I a mile off?
Click to expand...
You are more than a mile off. You simply do not have any real understanding of limits.
You need sit-down, face-to-face instruction. We here are not prepared to do that for you.

Let me tell you that I have taught about limits since 1964: high school honor mathematics, university mathematics, and graduate level analysis. I have taught from at least 100 textbooks and/or authors; but I have never seen anything like the rules you have been burdened with. If I were you, I would complain about this as incompetence.

There are some fundamental limits that all of this is based upon. You need to understand basic ideas.
\(\displaystyle \displaystyle{\lim _{n \to \infty }}\frac{c}{n} = 0\), that is you given reciprocal rule.

If \(\displaystyle |a|<1\) then \(\displaystyle \displaystyle{\lim _{n \to \infty }}a^n = 0\), that is your multiplication rule.

Example: \(\displaystyle \displaystyle{\lim _{n \to \infty }}\frac{{3n + 1}}{{{{(.8)}^n}n - 2n}} = {\lim _{n \to \infty }}\frac{{3 + \frac{1}{n}}}{{{{(.8)}^n} - 2}} = - \frac{3}{2}\)
 
pka said:
You are more than a mile off. You simply do not have any real understanding of limits.
You need sit-down, face-to-face instruction. We here are not prepared to do that for you.

Let me tell you that I have taught about limits since 1964: high school honor mathematics, university mathematics, and graduate level analysis. I have taught from at least 100 textbooks and/or authors; but I have never seen anything like the rules you have been burdened with. If I were you, I would complain about this as incompetence.

There are some fundamental limits that all of this is based upon. You need to understand basic ideas.
\(\displaystyle \displaystyle{\lim _{n \to \infty }}\frac{c}{n} = 0\), that is you given reciprocal rule.

If \(\displaystyle |a|<1\) then \(\displaystyle \displaystyle{\lim _{n \to \infty }}a^n = 0\), that is your multiplication rule.

Example: \(\displaystyle \displaystyle{\lim _{n \to \infty }}\frac{{3n + 1}}{{{{(.8)}^n}n - 2n}} = {\lim _{n \to \infty }}\frac{{3 + \frac{1}{n}}}{{{{(.8)}^n} - 2}} = - \frac{3}{2}\)
Click to expand...

For what it is worth pka, I agree with you. What I have noticed in my experience over recent years in our country is a attitude of mind that seems to have built up whereby instead of teaching by example and putting some effort into helping the learner understand, the material is written in a style that is testing your ability to solve problems before you have been educated in the subject itself. It seems that there is no foundation training to provide the basic skills, and while I don't want to condemn every written maths book, I have found it difficult to find books that actually learn you the basics, they just test you.

Take these sequences now as an example, if the written material I have to date, if the author had put some effort into showing by example how to work with these sequences from the beginning, then I would not be asking for help because the material would have provided it, but it doesn't.

Here is what is supposed to be a work example.

a_n = (60) / (3/n + 5)

The worked solution.

Now (3/n) converges to 0 for large n, so (3/n + 5) converges to 5. As n increases, a_n becomes arbitrary close to (60/5) = 12.

We conclude that a_n converges and that the lim n tends to infinity a_n = 12

What do you think about this type of course work, and do you agree that the learner can or cannot learn from that type of material presentation?
 
Probability said:
OK I'll try again.

a_n = (12 - 8n) / (4n + 36) (n = 1,2,3...)

I am looking at trying to understand convergence with these types of sequences and their limits. Given that this is a brand new topic to me I am blinded with information that is OTT for a beginner.

I assume that the test for convergence means that a_n tends to a definite value as n tends to infinity.

I am assuming that the sequence is n = 1,2,3..., therefore n will increase.

If the other member who posted is correct (I have little understanding at the moment) then;

12 - 8 = 4n, and

4 + 36 = 40n, therefore

4n / 40n

In which case I assume based on the product rule and the sequence n = 1.2.3..., that this sequence will not converge and will not tend to 0.

Please advise.

Thanks
Click to expand...

I just noticed that, in this post, you speculate that 12 - 8 = 4n (from 12 - 8n), and that 4 + 36 = 40n (from 4n + 36). This is NOT correct, and indicates a lack of understanding of basic algebra. If you are not familiar with things like combining like terms, then you are way over your head in trying to deal with understanding convergence.

Please confer with your academic advisor about whether you are properly placed in your currrent math class.
 
Reference

Paul's Online Math Notes

look up Calculus II, his notes on series and sequences are followed by many detailed examples of how to find the limit value of many types of expressions.

http://tutorial.math.lamar.edu/

He is very readable.
 
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