Verified answer

Isn't this just the Cauchy-Schwarz inequality?

Steve

Discrete math is really not my thing, but here's what I have so far.

Let ∑ (a_n)² = A

and ∑ (b_n)² = B

where A, B are real finite numbers

Take the product of the two equations:

{∑ (a_n)² } {∑ (b_n)² } = AB = real finite as well

LHS = a_0² b_0² + a_0² b_1² + ...

> (a_0 b_0)² + (a_1 b_1)² + ...

(*)

i.e. ∑ (a_n * b_n)² < AB

or ∑ (c_n)² < AB

and thus converges (absolutely)

All that's left is to show that if ∑ (c_n)² converges, then ∑ |c_n| converges.

(*) This line might be affected if a_n and b_n can take complex values, because then a_n² and b_n² will not be all positive.

This sounds like a job for ksoileau and jered.

*EDIT*

OK then. I looked up the Cauchy Schwarz inequality.

http://en.wikipedia.org/wiki/Cauchy-Schwarz_inequa...

You learn something new everyday.

Since this is already answered, I'll just point out to Dr.D:

It is FALSE that ∑An² converges ⇒ ∑|An| converges, as you would have figured out with a moment's reflection. :-)

E.g., consider An=1/n...!

In fact, the converse is true: ∑|An| converges ⇒ ∑An² converges absolutely! This is because

∑|An| converges ⇒An→0

⇒ |An|² < |An| for sufficiently large n, and so ∑|An|² converges by comparison, whence ∑An² converges absolutely.

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