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

by math25
Tags: convergent, counter, nonn, prove, series
 P: 25 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
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 Quote by 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?
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 $\sum a_i b_i$?
P: 25
 Quote by jbunniii 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

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

 Quote by math25 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?
 P: 25 the Cauchy-Schwarz inequality: sum ai bi is less then or equal to sum (ai^2)^1/2 sum (bi^2)^1/2
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 Quote by math25 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?
 P: 25 if ai converges (finite) then (ai)2 converges also.
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 Quote by math25 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?
 P: 25 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
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 Quote by 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
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 $x^2 \leq x$?
 Sci Advisor HW Helper Thanks P: 25,228 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.