Absolute convergence of series?

SMA_01
Messages
215
Reaction score
0
The question is: Show that if \suman from n=1 to ∞ converges absolutely, then \suman2 from n=1 to converges absolutely.

I'm not sure which approach to take with this.

I am thinking that since Ʃan converges absolutely, |an| can be either -an or an and for Ʃan2, an can be either negative or positive. But I'm not sure where to go with this.

Any help is appreciated.

Thank you.
 
Physics news on Phys.org
You know that if a_n is small (=smaller than 1), then a_n^2 is even smaller. So the squared series is smaller than the original series.

Can you do something with this idea??
 
So you mean a_n^2 would be bounded from above for smaller values? But what if it's greater than 1?
 
SMA_01 said:
So you mean a_n^2 would be bounded from above for smaller values? But what if it's greater than 1?

Yes, if it's greater than 1 then there is a problem. Try to find a way to solve this.
 
Do you think I can use what I stated in my first post?
 
SMA_01 said:
Do you think I can use what I stated in my first post?

Uh, not really. Well,it's not wrong what you said, but it won't help you much further.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top