Jump to content

absolute convergence


Sarahisme

Recommended Posts

The inequality holds; though it should be "strictly less than", unless all the terms are 0. But you can not just "make it up" and hope it's true, so understand why it is true.

Link to comment
Share on other sites

well when i say "made it up" i ment i sorta tryed a few cases, and saw that it worked, so then assumed it worked for all cases (quite dodgey i know).

 

would you be able to point me to somewhere where i can find this inequality written out properly (i.e. a maths webpage or something?)

 

also i am gathering if that ineqaulity holds, then the rest of the proof is correct?

 

Thanks for the reply DQW:)

Link to comment
Share on other sites

would you be able to point me to somewhere where i can find this inequality written out properly (i.e. a maths webpage or something?)
I doubt you'd find it written out anywhere. All it takes to prove it, is writing out explicitly, each term of the expression on either side of the inequality and multiplying out the terms on the RHS.

[math]eg:~~\sum_n |a_n| = (|a_1|~+~|a_2|~+~|a_3|~+~\dots ) [/math]

Link to comment
Share on other sites

The inequality holds; though it should be "strictly less than", unless all the terms are 0. But you can not just "make it up" and hope it's true, so understand why it is true.

 

i am not sure this is true... (i'm probably wrong, so correct if i am :P )

 

well what if the sum just goes from n = 1 to 1 ??

 

then it would be <=

 

wouldn't it? or ....not?

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.