Is it necessary to have two sequences for convergence or divergence to occur?

[pairofstrings]

Homework Statement

A bounded sequence need not be convergent

Can you show me an illustration which shows a sequence that is convergent?
I don't understand when if lim n ---> infinity s_n = l, the sequence s_n converges to l or {s_n}. Now what is l ?

My attempt or understanding of the convergence of sequence or series is similar to that of curve tracing. which I think is wrong. please tell me why are we performing root's test or ratio test and other test on the sequence? If it's to find whether or not the series is convergent then please explain what convergence is. Please tell me how curve tracing is different from convergence and divergence. I am getting confused between curve tracing and convergence.

 

[lswtech]

(-1)^n

 

[pairofstrings]

(-1)^n

What does it mean when the above series is said to converge. Can you draw me a figure?

 

[lswtech]

A bounded sequence need not be convergent : (-1)^n is an example - it is bounded within 1 and -1 ( abs((-1)^n))<1 ) while it is not convergent.

Series is just summation of sequences, say , Sn = 1+1/2^2+1/3^2+...+1/n^2 . Say, in fact, lim n-> infinity Sn = pi^2/6.

When we say something to be converge, your understanding using a curve sketching is OK. Consider 1/n. It is a sequence (not a series as series is summation of sequences!) and with n getting bigger and bigger, it turns closer and closer to 0. I don't know whether you have heard the epsilon-delta proofs, and it would be more formal to describe the things above.

But pay attention that Series/Sequence are just taking their values at integers and when we are talking about limits, we are talking about positive infinity only while Functions could take on any values.

Wish you love maths!

 

[pairofstrings]

A bounded sequence need not be convergent : (-1)^n is an example - it is bounded within 1 and -1 ( abs((-1)^n))<1 ) while it is not convergent.

Series is just summation of sequences, say , Sn = 1+1/2^2+1/3^2+...+1/n^2 . Say, in fact, lim n-> infinity Sn = pi^2/6.

When we say something to be converge, your understanding using a curve sketching is OK. Consider 1/n. It is a sequence (not a series as series is summation of sequences!) and with n getting bigger and bigger, it turns closer and closer to 0. I don't know whether you have heard the epsilon-delta proofs, and it would be more formal to describe the things above.

But pay attention that Series/Sequence are just taking their values at integers and when we are talking about limits, we are talking about positive infinity only while Functions could take on any values.

Wish you love maths!

I have attached a picture from wikipedia on sequences.
Now, my understanding so far, if correct, is that the summation of these sequences (blue dots) with another sequence is a series. Now, my question is how is this resultant series is said to be converging? Can you please draw me a curve?

 

[lswtech]

oh yes I understand your question just now.
The pic you attached from Wiki is an example of simple sequences only.

Pay attention to the fact that, a series is also a sequence. The same rule applys. Technically, when we are talking about "convergence", then we need to have some real L that lim n->inf Sn=L

Example of series that are convergent: 1/1+1/2!+1/3!+1/4!+...+1/n!
and 1/1^3+1/2^3+1/3^3+...+1/n^3

The above two series are the sum of 1/n! and 1/n^3 respectively.

Example of series that are NOT convergent: 1+1+1+1+1+...
and 1/1+1/2+1/3+...+1/n

Reason for 1+1+1+... is intuitive and for the second, let it to be Sn. Then S2n-Sn>1/2. It contradicts limS2n=limSn and thus limit does not exist.

Just to be simple - Series are sequences in nature!

 

[pairofstrings]

Example of series that are convergent: 1/1+1/2!+1/3!+1/4!+...+1/n!
and 1/1^3+1/2^3+1/3^3+...+1/n^3

The above two series are the sum of 1/n! and 1/n^3 respectively.

Example of series that are NOT convergent: 1+1+1+1+1+...
and 1/1+1/2+1/3+...+1/n

Just to be simple - Series are sequences in nature!

Can you explain to me the meaning of convergence and divergence from mathematical point of view?

I want to know if it's compulsory to have two sequences for them to converge or diverge.
Tell me if series also converges/diverges or not?
When do you call it that the sequence is converging or diverging?
What happens when series/sequence converges/divergence? Can you show me a diagram?

 

[Office_Shredder]

Can you explain to me the meaning of convergence and divergence from mathematical point of view?

I want to know if it's compulsory to have two sequences for them to converge or diverge.

It doesn't make sense to talk about the convergence or divergence of a pair of sequences really. Each sequence on its own either converges or diverges.

Given a list of numbers, you want to know if the numbers get really close to a single number. For example, the list [itex]1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},...[/itex] gets really close to 0. You may also think that it gets really close to .01 as well. The key to a sequence converging to a number is that it can get as close as you want to the number it converges to. For example, I might demand that the sequence be within .00001 of the number it converges to. Well, as long as I start picking numbers like [itex] \frac{1}{100000}[/itex] or [itex]\frac{1}{100001}[/itex], the sequence is always within .00001 of 0. But it isn't within .00001 of .01, and no matter how much farther along the sequence I go, I can never find numbers that come within .00001 of .01. So the sequence does NOT converge to .01. It does converge to 0 though, because no matter how small of a number you give me, I can find a point along the sequence after which the numbers in the sequence are always that close to 0.

Tell me if series also converges/diverges or not?

Everytime you see a series, you're just looking at a sequence in disguise. The series [itex] \sum_{n=0}^{\infty} \frac{1}{2}^n = 1+\frac{1}{2}+\frac{1}{4}+...[/itex] is really just the sequence [itex]1, \frac{3}{2}, \frac{7}{4},...[/itex]. You say the series converges if its associated sequence does.