Proving convergent for the products of two series.

[Disowned]

Homework Statement

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 \sumA_n and \sum B _n are both convergent series each with positive terms.
Show that the series \sumA_n B_n 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

The Attempt at a Solution

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

f(x) = \sumA_n
g(x) = \sumB_n

Known:

\sumA_n for all values
\sumB_n 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) = \sumA_nB_n

= 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 \sumA_n and \sumB_n convergent it follows that \sumA_nB_n
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?

 

[HallsofIvy]

Homework Statement

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 \sumA_n and \sum B _n are both convergent series each with positive terms.
Show that the series \sumA_n B_n 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

The Attempt at a Solution

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

f(x) = \sumA_n
g(x) = \sumB_n

Known:

\sumA_n for all values
\sumB_n 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) = \sumA_nB_n

= 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 \sumA_n and \sumB_n convergent it follows that \sumA_nB_n
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?

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"?

 

[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?

 

[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 A_kB_r is always positive.

 

[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?

 

[Disowned]

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?

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

This is what I wrote:

Given:
\sumA_n converges and \sumB_n converges
A_n is positive, B_n is positive.

Additional Info
If \sumA_n converges then the sequence A_n \stackrel{}{\rightarrow}0, for some N.

If this is true n > N.

Comparison Test Theorem.
If \sumB_n converges and A_n \leqB_n for all n, then \sumA_n converges also.

By definition one could infer that because A_n \leqB_n and A_n < 1 for some N then that must mean that B_n <1 for some N as well. Thus A_nBn_n < 1. Thus we have \sumA_nB_n is convergent by comparison.

That should be correct.

 

[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&lt;\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).

 

[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 A_nB_n < B_n but > A_n?

To clarify:

[ -1 <= A_n < A_nB_n < B_n < 1]?

 

[Dick]

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 A_nB_n < B_n but > A_n?

To clarify:

[ -1 <= A_n < A_nB_n < B_n < 1]?

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.

 

[Disowned]

I see, thanks for your help.