# Proving convergent for the products of two series.

1. Aug 17, 2008

### Disowned

1. The problem statement, all variables and given/known data

I'm been trying to wrap my head around this one for a couple of days. I have a problem that states the following.
(All sums go from n to $$\infty$$" Suppose we know that $$\sum$$An and $$\sum$$ B n are both convergent series each with positive terms.
Show that the series $$\sum$$An Bn must be convergent. I thought that if I took the first three terms of the series and multiply them together I could argue that since the terms alone are convergent then the product must be as well. However I'm not sure of that is the correct way to go about it.

Maybe I should take the derivative of each product and then make that argument. Anyone

3. The attempt at a solution

Here's what I wrote down while trying to solve this problem.

f(x) = $$\sum$$An
g(x) = $$\sum$$Bn

Known:

$$\sum$$An for all values
$$\sum$$Bn for all values

f(0) + f(1) + f(2) = A0 + A1 + A2
g(0) + g(1) + g(2) = B0 + B1 + B2

f(x) * g(x) = $$\sum$$AnBn

= A0 + A1 + A2(B0 + B1 + B2
)

[From here I foil the expression, which is a nightmare type unless you type at 120 wpm. So I'm going to skip that part.]

Since all values of $$\sum$$An and $$\sum$$Bn convergent it follows that $$\sum$$AnBn
is convergent as well.

Since I know that's not going to fly given there isn't all real proof shown, what else could I do? Do I take a derivative as mentioned or can I express the two sums in their elementary state?

2. Aug 17, 2008

### HallsofIvy

Staff Emeritus
But $(A_1+ A_2+ A_3+ \cdot\cdot\cdot)(B_1+ B_2+ B_3+ \cdot\cdot\cdot$
is not at all $\sum A_nB_n$. That would be simply $A_1B_1+ A_2B_2+ A_3B_3+ \cdot\cdot\cdot$

And you last paragraph seems to be just asserting that "Since the two sums are separately convergent, the product is"- which is exactly what you are supposed to prove!
What do you mean by "all values"?

3. Aug 17, 2008

### Disowned

That's the point I was trying to make. I didn't think I answered the question at all. By all values I meant for every value of the sum.

So now I know that wasn't the right solution, where do I go from here?

4. Aug 17, 2008

### Defennder

You should use an apparent inequality here. You're already told that An and Bn are always positive. What is the expression for f(x)*g(x)? Write it out using double summation symbols and take note that each term for AkBr is always positive.

5. Aug 17, 2008

### Dick

This is a LOT easier than than you think it is. If An is convergent, then the sequence An->0. So for some N, An<1 for n>N. Doesn't that suggest you use a comparison test with the sum of the Bn?

6. Aug 18, 2008

### Disowned

Thanks for the tip. Based on what you said I think I got a handle on the problem.

This is what I wrote:

Given:
$$\sum$$An converges and $$\sum$$Bn converges
An is positive, Bn is positive.

If $$\sum$$An converges then the sequence An $$\stackrel{}{\rightarrow}$$0, for some N.

If this is true n > N.

Comparison Test Theorem.
If $$\sum$$Bn converges and An $$\leq$$Bn for all n, then $$\sum$$An converges also.

By definition one could infer that because An $$\leq$$Bn and An < 1 for some N then that must mean that Bn <1 for some N as well. Thus AnBnn < 1. Thus we have $$\sum$$AnBn is convergent by comparison.

That should be correct.

7. Aug 18, 2008

### Dick

That's a little garbled. If N is such that An<1 for n>N, then $\sum_{n=N}^\infty A_n B_n<\sum_{n=N}^\infty B_n$. That's the essential fact. It works because i) An->0 and ii) you can discard any finite number of terms in a comparison test (since the sum of the terms you've discarded is finite).

8. Aug 18, 2008

### Disowned

Ah, I didn't think of that. So, in a sense, I'm really just using the squeeze theorem to prove convergence then? Where convergence is proven due to AnBn < Bn but > An?

To clarify:

[ -1 <= An < AnBn < Bn < 1]?

9. Aug 18, 2008

### Dick

Everything is positive so the lower bound is zero. All you need is 0<An*Bn<Bn. That's enough of a squeeze to show An*Bn converges.

10. Aug 18, 2008

### Disowned

I see, thanks for your help.