Series convergence- why is my proof incorrect?

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Homework Help Overview

The discussion revolves around proving the convergence of the series \(\sum a_{n}b_{n}\) given that \(\sum |a_{n}|\) converges and \((b_{n})\) is a bounded sequence. The original poster presents a proof using the Comparison Test and seeks clarification on the feedback received from their professor regarding the proof's validity.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use the Comparison Test to establish the convergence of \(\sum a_{n}b_{n}\) based on the convergence of \(\sum |a_{n}|\) and the boundedness of \((b_{n})\). Some participants question the application of the Comparison Test, particularly regarding the distinction between absolute and conditional convergence.

Discussion Status

Participants are actively discussing the original proof and the professor's feedback. There is a recognition of the need to clarify the relationship between absolute convergence and the conditions provided in the problem. Some participants express uncertainty about the necessity of using the Cauchy criterion as suggested by the professor.

Contextual Notes

The original poster notes that the problem statement included a hint to use Theorem 14.4, which states that a series converges if it satisfies the Cauchy Criterion. There is some debate about the implications of this hint and whether it was a requirement for the proof.

JThompson
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Homework Statement



Prove that if [itex]\sum{|a_{n}|}[/itex] converges and [itex](b_{n})[/itex] is a bounded sequence, then [itex]\sum a_{n}b_{n}[/itex] converges.

Homework Equations



Comparison Test part (i): Let [itex]\sum a_{n}[/itex] be a series where [itex]a_{n}\geq 0[/itex] for all n. If [itex]\sum a_{n}[/itex] converges and [itex]|b_{n}|\leq a_{n}[/itex] for all n, then [itex]\sum b_{n}[/itex] converges.

or Cauchy criterion

The Attempt at a Solution



Upon seeing this problem, I immediately thought of a proof using the Comparison Test which seemed easier that using the Cauchy criterion.
Proof:

Since [itex](b_{n})[/itex] is bounded, [itex]\exists M\in\mathbb{R}[/itex] with [itex]|b_{n}|\leq M[/itex] for all n. Then [itex]|a_{n}b_{n}|\leq M|a_{n}|[/itex].
Since [itex]\sum |a_{n}|[/itex] converges, [itex]\sum M|a_{n}|[/itex] converges (we proved this previously).
Since [itex]|a_{n}b_{n}|\leq M|a_{n}|[/itex] and [itex]\sum M|a_{n}|[/itex] converges, [itex]\sum a_{n}b_{n}[/itex] converges by the Comparison Test.

There were no marks indicating which part was incorrect, but I received half credit for the problem and my professor's comment was, "[itex]\sum |a_{n}b_{n}|[/itex] converges, use Cauchy criterion." Where does this proof run afoul?
 
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As your professor's note indicates, you showed that Σ |anbn| converges. You didn't show Σ anbn converges yet.
 
But according to the Comparison Test

JThompson said:
Comparison Test part (i): Let [itex]\sum a_{n}[/itex] be a series where [itex]a_{n}\geq 0[/itex] for all n. If [itex]\sum a_{n}[/itex] converges and [itex]|b_{n}|\leq a_{n}[/itex] for all n, then [itex]\sum b_{n}[/itex] converges.
[itex]M|a_{n}|\geq 0[/itex] is obvious since [itex]M\geq 0[/itex]. I proved that [itex]\sum M|a_{n}|[/itex] converges and that [itex]|a_{n}b_{n}|\leq M|a_{n}|[/itex]. It seems to me by the Comparison Test above that [itex]\sum a_{n}b_{n}[/itex]. The final series in the Comparison Test is not an absolute value, so why can I not do the same in this problem? How am I applying the Comparison Test incorrectly?
 
Yeah, you're right. Perhaps someone else can spot what we're both missing, or maybe your professor is wrong. You could go ask for clarification.
 
I'll ask him if no one responds- it's not urgent- but I have an easier time articulating questions online (because I'm shy), and written responses are easier to understand than verbal responses.
 
vela said:
As your professor's note indicates, you showed that Σ |anbn| converges. You didn't show Σ anbn converges yet.
Yes he did.
If

[tex] \sum |a_nb_n|[/tex]

then

[tex]\sum a_n b_n[/tex]

converges (absolute convergence)
 
statdad said:
Yes he did.
If

[tex] \sum |a_nb_n|[/tex]

then

[tex]\sum a_n b_n[/tex]

converges (absolute convergence)
D'oh! I hate it when I miss obvious stuff like that.

I've asked for others to take a look at this thread to see why the professor didn't like the proof.
 
Did the problem statement tell you to use the Cauchy criteria?
 
The only part of the problem statement that I did not post above was the book's hint.

Hint: Use Theorem 14.4

Theorem 14.4 states that a series converges iff it satisfies the Cauchy Criterion.
A hint is a suggestion, not a requirement. At least, I assumed as much.
 
  • #10
I see nothing wrong with your proof. Using the Cauchy criterion would only obfuscate it, in my opinion.
 

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