I can't seem to understand the ratio test proof

In summary: But thank you for the clarification! The proof makes much more sense now.In summary, the ratio test proof involves using the definition of the limit of a sequence to show that if |an+1/an| < 1, then the series \sum_{k=1}^{\infty}|a_{k}| converges. This is done by choosing a large enough N so that the terms of the series are bounded by a convergent geometric series.
  • #1
A.MHF
26
1
Hi everyone, I'm currently taking Calc II course and I'm kind of stuck in this ratio test proof thing.

Homework Statement



http://blogs.ubc.ca/infiniteseriesmodule/appendices/proof-of-the-ratio-test/proof-of-the-ratio-test/

I'm trying to understand the proof, but there are some parts that I don't really get.
So assuming that |an+1/an| = L < 1
and that L < r
|an+1/an| < r
but then, the proof says there is an integer N which is < or = to n. From where did that come from? Why do we need it in the first place? And how does it relate to the proof?
The proof goes on and I don't really have any idea what's going on.

Homework Equations


-

The Attempt at a Solution


I tried making a number line to understand what's going on but this seems so subtle.
 
Physics news on Phys.org
  • #2
there are some small typos in the proof which maybe make it harder to understand it, but anyway that N comes from the definition of the limit of a sequence (this is stated in the proof that it follows from the formal definition of the limit). Intuitively when we know that a sequence Cn has limit L this means that Cn gets closer to L as n grows larger. So if we want Cn to be close enough to L so that it is L<Cn<r all we have to do is to choose a large enough N and for all n>=N it will be L<Cn<r.

The proof utilizes of this fact for the sequence Cn=|an+1/an|. It uses this N to prove that |aN+k|<|aN|rk. From this last inequality you can prove that the series [itex]\sum_{k=1}^{\infty}|a_{N+k}|[/itex] converges (because it is bounded by the geometric series [itex]\sum_{k=1}^{\infty}|a_{N}|r^k[/itex] which obviously converges). Now the series [itex]\sum_{k=1}^{\infty}|a_{k}|[/itex] also converges because it differs from the [itex]\sum_{k=1}^{\infty}|a_{N+k}|[/itex] only by a finite sum [itex]\sum_{k=1}^{N}|a_k|[/itex]
 
Last edited:
  • Like
Likes A.MHF
  • #3
A.MHF said:
Hi everyone, I'm currently taking Calc II course and I'm kind of stuck in this ratio test proof thing.

Homework Statement



http://blogs.ubc.ca/infiniteseriesmodule/appendices/proof-of-the-ratio-test/proof-of-the-ratio-test/

I'm trying to understand the proof, but there are some parts that I don't really get.
So assuming that |an+1/an| = L < 1
The proof doesn't say ##\lvert a_{n+1}/a_n \rvert = L##. It says that
$$\lim_{n \to \infty} \left\lvert \frac{a_{n+1}}{a_n} \right\rvert = L.$$ If you understand what the limit means, the proof should make more sense to you.
 
  • #4
Delta² said:
there are some small typos in the proof which maybe make it harder to understand it, but anyway that N comes from the definition of the limit of a sequence (this is stated in the proof that it follows from the formal definition of the limit). Intuitively when we know that a sequence Cn has limit L this means that Cn gets closer to L as n grows larger. So if we want Cn to be close enough to L so that it is L<Cn<r all we have to do is to choose a large enough N and for all n>=N it will be L<Cn<r.

The proof utilizes of this fact for the sequence Cn=|an+1/an|. It uses this N to prove that |aN+k|<|aN|rk. From this last inequality you can prove that the series [itex]\sum_{k=1}^{\infty}|a_{N+k}|[/itex] converges (because it is bounded by the geometric series [itex]\sum_{k=1}^{\infty}|a_{N}|r^k[/itex] which obviously converges). Now the series [itex]\sum_{k=1}^{\infty}|a_{k}|[/itex] also converges because it differs from the [itex]\sum_{k=1}^{\infty}|a_{N+k}|[/itex] only by a finite sum [itex]\sum_{k=1}^{N}|a_k|[/itex]

Thanks, that makes much more sense now!

Vela: yea sorry, I knew about the limit I just forgot to type it.
 

What is the ratio test?

The ratio test is a mathematical test used to determine the convergence or divergence of an infinite series. It compares the ratio of consecutive terms in the series to a limit, and if the limit is less than 1, the series is convergent. If the limit is greater than 1, the series is divergent.

How is the ratio test used to prove the convergence or divergence of a series?

The ratio test is used to prove the convergence or divergence of a series by showing that the limit of the ratio of consecutive terms is less than or greater than 1. If the limit is less than 1, the series is convergent, and if the limit is greater than 1, the series is divergent.

What is the proof for the ratio test?

The proof for the ratio test involves using the limit comparison test and the definition of convergence to show that the limit of the ratio of consecutive terms is less than or greater than 1. This is done by manipulating the series and using algebraic properties to simplify the expression until the limit can be evaluated.

What are the limitations of the ratio test?

The ratio test can only be used for series with positive terms, and it may not always give a conclusive result. In certain cases, the limit of the ratio may be equal to 1, which means the test is inconclusive and another test, such as the root test, must be used.

How can I improve my understanding of the ratio test proof?

To improve your understanding of the ratio test proof, it is important to have a solid understanding of limits, series, and algebraic manipulation. It can also be helpful to work through example problems and practice applying the ratio test to different series. Seeking help from a math tutor or attending review sessions can also aid in improving your understanding.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
711
  • Calculus and Beyond Homework Help
Replies
2
Views
186
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
24
Views
795
  • Calculus and Beyond Homework Help
Replies
1
Views
255
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
842
  • Calculus and Beyond Homework Help
Replies
2
Views
602
Replies
6
Views
678
Replies
23
Views
2K
Back
Top