Analysis Proof (simple, but )

[futurebird]

Analysis Proof (simple, but urgent!)

Homework Statement

This is an exercise from Rudin Chapter 3.

#8. If \sum a_{n} converges, and if {bn} is monotonic and bounded, prove that \sum b_{n} a_{n} converges.

Homework Equations

THEOREM: Suppose {sn} is monotonic. Then {sn} converges <==> {sn} is bounded.

The Attempt at a Solution

{bn} converges. bn --> b
So, for any \epsilon &gt; 0 \exists N such that n > N implies |b - b_{n}| &lt; \epsilon

Let M = b if b > bN
Let M = \epsilon + b if b <= bN

Now, \sum M a_{n} = M \sum a_{n} for M constant. And \sum M a_{n} converges to Ma. So, \sum b_{n} a_{n} converges.


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Am I on the right track with this? I need to turn this in soon and any help would be GREAT!

 

[dynamicsolo]

Maybe one of the mathematicians here will have more to say about this, but it strikes me that there are two cases to consider, since the proposition only says {b_n} is monotonic and bounded.

What you've treated is the case where {b_n} is monotonically increasing, so since b_n never gets bigger than a constant M, and the sum of M·a_n will converge, so will the sum of b_n · a_n .

You may also need to look at the case where {b_n} is monotonically decreasing. The basic argument is similar, but one other remark will need to be included...

 

[jostpuur]

I didn't fully understand what you were doing, but it looks like that you are trying to use the knowledge that the sequence b_n is converging. You should not try to prove this claim:

"Suppose that the series \sum_n a_n converges, and that the sequence b_n converges. Then the series \sum_n b_n a_n converges."

That is not right: Set

<br /> a_n = \frac{(-1)^n}{\sqrt{n}}<br />

<br /> b_{2n+1} = 0,\quad b_{2n}=\frac{1}{\sqrt{n}}.<br />

for a counter example.

Sorry, I don't know how to do that exercise, and cannot give more advice now, but I know that you must use the knowledge that b_n is monotonic and bounded for something more than the convergence of b_n.

 

[futurebird]

Thank you! That was helpful, I think I have a better proof now!

 

[dynamicsolo]

<br /> b_{2n+1} = 0,\quad b_{2n}=\frac{1}{\sqrt{n}}.<br />

Isn't b_{n} supposed to be monotonic for this proposition?

 

[futurebird]

That was a counter example, I think. Showing that bn being monotonic matters.

 

[konthelion]

#8. If \sum a_{n} converges, and if {bn} is monotonic and bounded, prove that \sum b_{n} a_{n} converges.

Consider two cases where b is a monotonely increasing sequence and monotonely decreasing.

Then you can use the Theorem 3.42 in Rudin's book to show that series \sum_n a_n \left(b_{n}-b\right) converges, then \sum_n a_n b_n converges, where b is the limit of the sequence b_{n}

Hint:\lim_{n \to \infty} \left(b_{n}-b\right) =0

 

[dynamicsolo]

That was a counter example, I think. Showing that bn being monotonic matters.

Right, sorry... On re-reading the entire thread, the context of some of the posts is clearer to me. I see the theorem jostpuur was remarking on. I was focused on the boundedness and thought that was what you were using in your argument. The theorem is not required for the proof.