Proving Convergence of Positive Series Using Ratio Test

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SUMMARY

The discussion focuses on proving the convergence of the series ∑an given that ∑bn converges and an/bn approaches 0. The key conclusion is that if ∑bn converges, then bn approaches 0, which implies that an must behave in a specific manner to ensure convergence. The ratio test is highlighted as a relevant method to establish this proof, emphasizing that the condition an/bn → 0 is a strong indicator that an is eventually smaller than bn.

PREREQUISITES
  • Understanding of series convergence, specifically the properties of convergent series.
  • Familiarity with the Ratio Test for series convergence.
  • Knowledge of limits and their implications in calculus.
  • Basic experience with formal proofs in mathematical analysis.
NEXT STEPS
  • Study the Ratio Test for series convergence in detail.
  • Explore the implications of the convergence of series and their terms approaching zero.
  • Review examples of series that converge and diverge to solidify understanding.
  • Practice proving convergence of series using various tests, including the Comparison Test.
USEFUL FOR

Students preparing for calculus or analysis exams, particularly those focusing on series convergence, as well as educators seeking to clarify the concepts of the Ratio Test and series behavior.

Jamin2112
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Homework Statement



I'm studying for my final next week (In other words, I'll be asking a lot of questions on here).

One practice problem from the book reads

Prove that, if ∑an and ∑bn are series of positive terms with ∑bn convergent and an/bn --> 0, then ∑an is convergent.

Homework Equations



Whatever

The Attempt at a Solution



So, if an/bn --> 0, then either

bn ---> ∞ or -∞ AND an ---> some finite number;

or bn ----> some finite number AND an ---> 0.

The second option is true given that ∑bn is convergent.


? I don't. Help me with a good way of explaining this.
 
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Take the following with a grain of salt, and perhaps ignore it as I don't have much experience with formalized dealings with series. But I have an intuitive suggestion that may lead to the proof. (Or it could be just down right wrong.)

If \Sigma b_{n} converges, I think that implies b_{n} goes to zero.
(The contrapositive of the theorem that if a sequence does not go to zero, then the series diverges.)
Now if b_{n} goes to zero, then this has some implications for what \frac{a_{n}}{b_{n}} actually means. Namely it would seem that if b_{n} is going to zero, then that would cause the fraction to go to infinity. Since it doesn't, this means a_{n} is forced to behave in a very particular way.

I think there lies the proof.
 
This is simple -- don't think too hard. The condition a_n/b_n \to 0 is actually much stronger than you need. You know \textstyle\sum b_n converges, and a_n/b_n \to 0 is a relatively strong way of saying that "a_n is eventually smaller than b_n". This should remind you of a particular test for series convergence which you can use to prove that \textstyle\sum a_n converges.
 

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