Prove: Sum ai bi Converges if Sum ai & Sum bi Non-negative

[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]

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

Correct. Can you find a counterexample? (Hint: you can find one with $a_i = b_i$.)

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.

Do you know any inequalities that involve $\sum a_i b_i$?

 

[math25]

Correct. Can you find a counterexample? (Hint: you can find one with $a_i = b_i$.)ai= (cos n pie)/squareroot (n)= bi this would work right?Do you know any inequalities that involve $\sum a_i b_i$?

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]

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

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

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

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]

the Cauchy-Schwarz inequality:


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

OK, good. Now, given that all the $a_i$ are nonnegative, what can you say about

$$\sum a_i^2$$

if you know that

$$\sum a_i$$

is finite?

 

[math25]

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

 

[jbunniii]

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

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]

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

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.

Ai, Bi, Ci are obviously strictly increasing (1) ,

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 $a_i$ and $b_i$ are non-negative:

$$\sum a_i b_i \leq \sqrt{\sum a_i^2} \sqrt{\sum b_i^2}$$

Therefore, if $\sum a_i^2$ and $\sum b_i^2$ are finite, then so is $\sum a_i b_i$.

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

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

 

[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]

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.

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.