Proof of Convergence: a_n>=0 and summation a_n

[fk378]

Homework Statement

Prove that if a_n>=0 and summation a_n converges, then summation (a_n)^2 also converges.

The Attempt at a Solution

(Note: When I say "lim" please assume the limit as n-->infinity). I just want it to be a little clearer to read)

If summation a_n converges, then lim(a_n)=0. If lim(a_n)=0, then lim(a_n)^2=0.
If summation (a_n)^2 diverges, then lim(a_n)^2 does not equal 0. But lim(a_n)^2=0, so summation (a_n)^2 must converge.

Can anyone let me know if this is a valid proof? I'm not sure how else to prove it otherwise...thank you.

 

[jhicks]

If summation (a_n)^2 diverges, then lim (n-->inf) (a_n)^2 does not equal 0. But lim (n-->inf) (a_n)^2=0, so summation (a_n)^2 must converge.

I'm having some trouble following those lines. What about $$a_{n}=\frac{1}{\sqrt{n}}$$?

 

[fk378]

Oh, that does go against my proof. I don't know how else to prove it then. Any suggestions?

(I also edited a bit of my original post so that it would be a bit easier to read, hopefully)

 

[jhicks]

I'm pretty sure you posted this a week ago as well. Dick provided a succinct proof. I'm not saying it's the only way, but it is definitely easy.

https://www.physicsforums.com/showthread.php?t=214061

 

[HallsofIvy]

If summation (a_n)^2 diverges, then lim(a_n)^2 does not equal 0.

This is definitely NOT true!

However, it is true that if $\sum a_n$ converges then $lim a_n= 0$.
For sufficiently large n, a_n< 1 and so a_n²< |a_n|.