Proving Convergence of Positive Series Using Ratio Test: A Guide

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In summary, the problem asks for a series that converges but has an infinite value for one term. If this series converges, then the value of the term after the infinite term will be 0.
  • #1
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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  • #2
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 [tex]\Sigma b_{n}[/tex] converges, I think that implies [tex]b_{n}[/tex] goes to zero.
(The contrapositive of the theorem that if a sequence does not go to zero, then the series diverges.)
Now if [tex]b_{n}[/tex] goes to zero, then this has some implications for what [tex]\frac{a_{n}}{b_{n}}[/tex] actually means. Namely it would seem that if [tex]b_{n}[/tex] is going to zero, then that would cause the fraction to go to infinity. Since it doesn't, this means [tex]a_{n}[/tex] is forced to behave in a very particular way.

I think there lies the proof.
 
  • #3
This is simple -- don't think too hard. The condition [tex]a_n/b_n \to 0[/tex] is actually much stronger than you need. You know [tex]\textstyle\sum b_n[/tex] converges, and [tex]a_n/b_n \to 0[/tex] is a relatively strong way of saying that "[tex]a_n[/tex] is eventually smaller than [tex]b_n[/tex]". This should remind you of a particular test for series convergence which you can use to prove that [tex]\textstyle\sum a_n[/tex] converges.
 

1. What is the ratio test and how does it work?

The ratio test is a method used to determine the convergence or divergence of a series. It involves taking the limit of the absolute value of the ratio between the (n+1)th term and the nth term of a series. If the limit is less than 1, the series converges, and if it is greater than 1, the series diverges.

2. How do you use the ratio test to prove convergence of a positive series?

To use the ratio test, you need to take the limit of the absolute value of the ratio between the (n+1)th term and the nth term of the series. If the limit is less than 1, the series converges. You can then use the result of the limit to show that the series is bounded above by a convergent geometric series, which proves convergence.

3. What is a positive series?

A positive series is a series where all the terms are positive. This is important when using the ratio test because it only applies to positive series. If a series has both positive and negative terms, the ratio test cannot be used to determine convergence.

4. Can the ratio test be used to prove divergence of a positive series?

Yes, the ratio test can also be used to prove divergence of a positive series. If the limit of the ratio between the (n+1)th term and the nth term is greater than 1, the series diverges. However, it is important to note that if the limit is equal to 1, the test is inconclusive and another method must be used to determine convergence or divergence.

5. Are there any limitations to using the ratio test to prove convergence of a positive series?

Yes, there are some limitations to using the ratio test. It can only be applied to series with positive terms and it may not always give a conclusive result. In some cases, the limit of the ratio may be equal to 1, making the test inconclusive. In these situations, other methods such as the comparison test or the integral test may need to be used.

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