calculs 1

[no123]

i need help with something..
prove or disprove:
if you multiply a series who converged conditionally with a series who converges absolutely you get a series who converges absolutely.
if it's not true,you need to give an example.
thank you

 

[Ishuda]

i need help with something..
prove or disprove:
if you add a series who converged conditionally to a series who converges absolutely you get a series who converges absolutely.
if it's not true,you need to give an example.
thank you

What thoughts do you have on this? As a backward sort of hint, although I'm not sure it would be acceptable as an example, what if each term of the absolutely convergent series were zero.

 

[no123]

What thoughts do you have on this? As a backward sort of hint, although I'm not sure it would be acceptable as an example, what if each term of the absolutely convergent series were zero.

ohh..i haven't thought of that.
do you have an example with a different series?which isnt a constant number?it's quite difficult.

 

[no123]

What thoughts do you have on this? As a backward sort of hint, although I'm not sure it would be acceptable as an example, what if each term of the absolutely convergent series were zero.

i made a mistake,you dont add them,you need to multiply them...

 

[Ishuda]

i made a mistake,you dont add them,you need to multiply them...

Lets ramble a bit for a hint: For convenience we will let the terms for the conditional convergent series be c_n, the absolutely convergent series be a_n, and their product p_n = c_n a_n. If you have a convergent series (conditional or absolute) what is the limit of the individual terms? That is what is the limit as n goes to infinity of c_n? Of a_n? Given that, then, eventually, what is the size of p_n compared to a_n?

 

[no123]

Lets ramble a bit for a hint: For convenience we will let the terms for the conditional convergent series be c_n, the absolutely convergent series be a_n, and their product p_n = c_n a_n. If you have a convergent series (conditional or absolute) what is the limit of the individual terms? That is what is the limit as n goes to infinity of c_n? Of a_n? Given that, then, eventually, what is the size of p_n compared to a_n?

it's the same size,isnt it?

 

[Ishuda]

it's the same size,isnt it?

If \(\displaystyle \Sigma c_n\) converges, what is the limit as n goes to infinity of c_n?

 

[no123]

If \(\displaystyle \Sigma c_n\) converges, what is the limit as n goes to infinity of c_n?

0?

 

[Ishuda]

0?

Zero is correct. So at some point c_n must get to be less than 1 and stay less than 1. At that point what is the size of |c_n a_n| compared to |a_n|? Is |c_n a_n| greater than |a_n|? Equal to |a_n|? Less than |a_n|? Having answered that, what does that mean about the convergence of \(\displaystyle \Sigma c_n a_n\)?

 

[no123]

Zero is correct. So at some point c_n must get to be less than 1 and stay less than 1. At that point what is the size of |c_n a_n| compared to |a_n|? Is |c_n a_n| greater than |a_n|? Equal to |a_n|? Less than |a_n|? Having answered that, what does that mean about the convergence of \(\displaystyle \Sigma c_n a_n\)?

less than |a_n|?
do you have an example which disprove the claim?i'd really like for you to share with me,i have an exam tomorrow..

 

[stapel]

do you have an example which disprove the claim? i'd really like for you to share with me,i have an exam tomorrow..

Then you need to learn how to think about this stuff on your own, since of course Ishuda won't be available to help on your exam. This is likely why Ishuda is giving so much time and care in helping you "walk through" this material.

Now: Let's hear your thoughts; let's see your efforts.

 

[no123]

Then you need to learn how to think about this stuff on your own, since of course Ishuda won't be available to help on your exam. This is likely why Ishuda is giving so much time and care in helping you "walk through" this material.

Now: Let's hear your thoughts; let's see your efforts.

i think i have an example,if it's true can you give me a different example?
((-1)^n)/n converges conditionally and the constant series 1 for instance converged absolutely so the multiplication does not converges absolutely because we get the same series we took as conditionally converges.
i want to see an example with a series who converges absolutely which isn't a constant number..please do you have one?

 

[Ishuda]

less than |a_n|?
do you have an example which disprove the claim?i'd really like for you to share with me,i have an exam tomorrow..

If, for some N, |c_n a_n| < |a_n| for all n>N, and \(\displaystyle \Sigma a_n\) converges absolutely, what does that say about \(\displaystyle \Sigma c_n a_n\) by the comparison test.

 

[no123]

If, for some N, |c_n a_n| < |a_n| for all n>N, and \(\displaystyle \Sigma a_n\) converges absolutely, what does that say about \(\displaystyle \Sigma c_n a_n\) by the comparison test.

that |c_n a_n| also converges,er go \(\displaystyle \Sigma c_n a_n\) converges absolutely.
so the claim was true all along?..if so,can you write it down as you explained me-c_{n 's limit is 0 so at some point he becomes less than 1 and than it's size is smaller than} a_{n and all that..}

 

[Ishuda]

that |c_n a_n| also converges,er go \(\displaystyle \Sigma c_n a_n\) converges absolutely.
so the claim was true all along?..if so,can you write it down as you explained me-c_{n 's limit is 0 so at some point he becomes less than 1 and than it's size is smaller than} a_{n and all that..}

No. So far, you have basically done nothing unless prompted. You will have to do your own prompting for your test and you can start now by going back over all the posts leading up to the answer and write it down yourself. I don't mind helping someone but I draw the line at writing out a proof so that they can turn it in as there own work and that now appears to be all you want.