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I can't seem to understand the ratio test proof

  • Thread starter A.MHF
  • Start date
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Hi everyone, I'm currently taking Calc II course and I'm kind of stuck in this ratio test proof thing.
1. The problem statement, all variables and given/known data

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.
2. Relevant equations
-
3. The attempt at a solution
I tried making a number line to understand what's going on but this seems so subtle.
 

Delta2

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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:

vela

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Hi everyone, I'm currently taking Calc II course and I'm kind of stuck in this ratio test proof thing.
1. The problem statement, all variables and given/known data

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.
 
26
1
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.
 

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