Proving convergent for the products of two series.

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SUMMARY

The discussion focuses on proving the convergence of the series \(\sum A_n B_n\) given that both \(\sum A_n\) and \(\sum B_n\) are convergent series with positive terms. The participants clarify that the product of the series cannot be directly inferred from the convergence of the individual series. Instead, they suggest using the Comparison Test Theorem, which states that if \(A_n \leq B_n\) for all \(n\) and \(\sum B_n\) converges, then \(\sum A_n\) also converges. The final conclusion is that \(\sum A_n B_n\) converges by establishing that \(0 < A_n B_n < B_n\).

PREREQUISITES
  • Understanding of series convergence, specifically the properties of convergent series.
  • Familiarity with the Comparison Test Theorem in series.
  • Knowledge of the Squeeze Theorem and its application in proving convergence.
  • Basic algebraic manipulation of series and summations.
NEXT STEPS
  • Study the Comparison Test Theorem in detail to understand its applications in series convergence.
  • Learn about the Squeeze Theorem and how it can be applied to series and sequences.
  • Explore examples of proving convergence for products of series in mathematical literature.
  • Investigate other convergence tests, such as the Ratio Test and Root Test, for a broader understanding of series analysis.
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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 \sumAn and \sum B n are both convergent series each with positive terms.
Show that the series \sumAn 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

The Attempt at a Solution



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

f(x) = \sumAn
g(x) = \sumBn

Known:

\sumAn for all values
\sumBn 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) = \sumAnBn

= 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 \sumAn and \sumBn convergent it follows that \sumAnBn
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?
 
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Disowned said:

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 \sumAn and \sum B n are both convergent series each with positive terms.
Show that the series \sumAn 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



The Attempt at a Solution



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

f(x) = \sumAn
g(x) = \sumBn

Known:

\sumAn for all values
\sumBn 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) = \sumAnBn

= 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 \sumAn and \sumBn convergent it follows that \sumAnBn
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"?
 
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?
 
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.
 
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?
 
Dick said:
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:
\sumAn converges and \sumBn converges
An is positive, Bn is positive.

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

If this is true n > N.

Comparison Test Theorem.
If \sumBn converges and An \leqBn for all n, then \sumAn converges also.

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

That should be correct.
 
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).
 
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]?
 
Disowned said:
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]?

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
I see, thanks for your help.
 

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