prove or give a counter example is sum ai and sum bi are convergent series with non-n

math25

Hi

Can someone please help me to
prove or give a counter example is sum ai and sum bi are convergent series with non-negative terms then sum aibi converges


I believe that if it doesn't say "non-negative terms" then this wouldn't be true. Am I correct?

Since each of two non-negative series converges then the series sum ai bi converges also. However I am not sure how to prove this.

thanks

jbunniii

Correct. Can you find a counterexample? (Hint: you can find one with [itex]a_i = b_i[/itex].)

Do you know any inequalities that involve [itex]\sum a_i b_i[/itex]?

math25

ai= (cos n pie)/squareroot (n)= bi this would work right?

I'm sorry I am not sure what you mean....i feel like this problem its very simple but for some reason I have such a hard time with it

jbunniii

Yes, that's the example I had in mind. Of course [itex]\cos(n\pi) = (-1)^n[/itex].

Do you know the Cauchy-Schwarz inequality?

math25

the Cauchy-Schwarz inequality:


sum ai bi is less then or equal to sum (ai^2)^1/2 sum (bi^2)^1/2

jbunniii

OK, good. Now, given that all the [itex]a_i[/itex] are nonnegative, what can you say about

[tex]\sum a_i^2[/tex]

if you know that

[tex]\sum a_i[/tex]

is finite?

math25

if ai converges (finite) then (ai)2 converges also.

jbunniii

Correct, but do you know how to prove it?

And can you see how to use this fact along with the Cauchy-Schwarz inequality to solve the problem?

math25

Since I am not sure how to use the Cauchy ineq. , how about something like this :

Let
Ai=Sum[i=0 to inf] ai
Bi=Sum[i= 0 to inf] bi
Ci=Sum[i=0 to inf] ai* bi
Ai, Bi, Ci are obviously strictly increasing (1) , because
Ai=A_(i-1)+a_n, similary Bi and Ci.
Let lim(n->inf)Ai=X, lim(n->inf)Bi=Y (because they converge).
Because they are strictly increasing,
=>Ai=X and Bi=Y for every i.
Ci=Ai*Bi, because
a1*b1+a2*b2+...an*bn<(a1+a2+..+an)*
*(b1+b2+..+bn)
From this, Ci=Ai*Bi=X*Y (2)
From (1) and (2) (monotonous and limited) Ci is convergent

jbunniii

These definitions don't make any sense. On the right hand side, you have summed over i = 0 to infinity. The result therefore does not depend on i.

Since your definition of Ai, Bi, Ci above doesn't actually depend on i, this statement also makes no sense. A constant can't be strictly increasing.

Let's look at the Cauchy-Schwarz inequality again, which looks like the following assuming that [itex]a_i[/itex] and [itex]b_i[/itex] are non-negative:

[tex]\sum a_i b_i \leq \sqrt{\sum a_i^2} \sqrt{\sum b_i^2}[/tex]

Therefore, if [itex]\sum a_i^2[/itex] and [itex]\sum b_i^2[/itex] are finite, then so is [itex]\sum a_i b_i[/itex].

We know that [itex]\sum a_i[/itex] and [itex]\sum b_i[/itex] are finite. If you can show that this implies that [itex]\sum a_i^2[/itex] and [itex]\sum b_i^2[/itex] are finite, then you're done. So focus on this step.

Here's a hint: if [itex]x[/itex] is a nonnegative real number, what has to be true of [itex]x[/itex] in order to have [itex]x^2 \leq x[/itex]?

Dick

Actually you don't really need Cauchy-Schwarz do you? If the series ai converges then ai approaches 0. So ai<1 for i large enough, doesn't it? Use a comparison test.

jbunniii

Yes, that's much better. When I see an inner product I always think "Cauchy-Schwarz" but in this case it meant that I didn't notice the more direct proof.