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

• math25
In summary: Cauchy-Schwarz inequality?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...Cauchy-Schwarz inequality?In summary, this conversation is about whether a series of non-negative terms converges or not. The person is not sure how to prove that the series sum ai bi does in fact converge. They ask for help from someone who knows about the Cauchy-Schwarz inequality, and is told that in order for the series to converge, both the a_i and the b_i must be finite. They are then
math25
Hi

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

math25 said:
Hi

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 $\sum a_i b_i$?

jbunniii said:
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

math25 said:
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?

the Cauchy-Schwarz inequality:

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

math25 said:
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?

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

math25 said:
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?

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

math25 said:
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 $x^2 \leq x$?

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.

Dick said:
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.

## 1. What is the meaning of "Sum ai bi converges"?

The phrase "Sum ai bi converges" means that the sum of the products of two sequences, ai and bi, converges to a finite value. In simpler terms, it means that the sum of the terms in the sequence of ai multiplied by the terms in the sequence of bi approaches a definite value, rather than increasing or decreasing without bound.

## 2. Why is it necessary for Sum ai and Sum bi to be non-negative?

It is necessary for Sum ai and Sum bi to be non-negative because if either of these two sums are negative, then the sum of the products of ai and bi would not converge. This is because multiplying a negative number by any other number will result in a negative product, which would cause the sum to decrease or increase without bound.

## 3. What does it mean for a sequence to converge?

A sequence converges when the terms in the sequence approach a definite value or limit as the number of terms increases. In other words, the terms in the sequence get closer and closer to a specific value, rather than increasing or decreasing without bound.

## 4. Can you provide an example of when Sum ai bi converges?

One example of when Sum ai bi converges is when ai and bi are both positive, and the two sequences are defined as follows: ai = 1/n and bi = n, where n is a positive integer. The sum of the products of these two sequences would be the sum of the series 1 + 2 + 3 + 4 + ..., which is known to converge to a finite value (in this case, the value is infinity).

## 5. How is this statement proven?

This statement can be proven using the Cauchy-Schwarz inequality, which states that for any two sequences of numbers, the sum of the products of the terms of these sequences is less than or equal to the product of the sums of the squares of these sequences. By applying this inequality to the sequences ai and bi, we can show that the sum of the products of these two sequences must converge if the sums of ai and bi are non-negative.

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