convergent series

[sambellamy]

I have a_n+1 = 1/(3-a_n). I am being asked to prove that it is convergent. I see from plugging in values that is converges on 0, and is decreasing. Am I supposed to use induction to do this? So far I have done the following:
decided that a_n = 1/(3-a_n+1)
tried to split up the definition into 1/3 - 1/a_n
tried to use a_n > a_n+1
but i am coming up short. Can i get a little guidance?

 

[stapel]

I have a_n+1 = 1/(3-a_n). I am being asked to prove that it is convergent.

What tests or methods have you been taught for this? You should probably use one of those.

 

[sambellamy]

Well that's just the issue - I'm looking for some pointers on which method to use. is induction the right way to go, and if so, am i approaching it in the right way? what would be the next step? I don't have a lot of practice with induction.

 

[stapel]

Well that's just the issue - I'm looking for some pointers on which method to use. is induction the right way to go, and if so, am i approaching it in the right way? what would be the next step? I don't have a lot of practice with induction.

Since apparently you've been taught and told no methods to you, you might as well use the standard sort of proof, which you can begin studying here.

 

[sambellamy]

So, is that a no vote for induction?

 

[HallsofIvy]

I have a_n+1 = 1/(3-a_n). I am being asked to prove that it is convergent. I see from plugging in values that is converges on 0, and is decreasing. Am I supposed to use induction to do this? So far I have done the following:
decided that a_n = 1/(3-a_n+1)
tried to split up the definition into 1/3 - 1/a_n

Well, that's a major mistake right there!
\(\displaystyle \frac{1}{3- a_n}\ne \frac{1}{3}- \frac{1}{a_n}\)

tried to use a_n > a_n+1
but i am coming up short. Can i get a little guidance?

One difficulty is that you haven't said what the first term in the series is! Also are you talking about the series \(\displaystyle \sum a_n\) or the sequence \(\displaystyle \{a_n\}\)? To prove that the series, \(\displaystyle \sum a_n\), converges, it must be true that the sequence \(\displaystyle \{a_n\}\) converges to 0 but the converse is not true- even if the sequence converges to 0, the series might not converge.

 

[sambellamy]

Thank you for your reply! The first term of the series is a₁ = 2. Also, I am only concerned with the sequence, not the sum/series.

 

[pka]

Thank you for your reply! The first term of the series is a₁ = 2. Also, I am only concerned with the sequence, not the sum/series.

Induction is so easy. Clearly \(\displaystyle a_1>a_2\).
So suppose \(\displaystyle a_K<a_{K-1}.\)
Then \(\displaystyle -a_K>-a_{K-1}\\3-a_K>3-a_{K-1}\\\dfrac{1}{3-a_K}< \dfrac{1}{3-a_{K-1}}\)

Can you finish?

 

[sambellamy]

Thank you for your reply. This is very helpful.

Induction is so easy. Clearly \(\displaystyle a_1>a_2\).
So suppose \(\displaystyle a_K>a_{K-1}.\)

If a₁ > a₂ wouldn't a_k > a_k+1 and a_k < a_k-1 ? I used these and got:
a_k-1 > a_k
-a_k-1 < -a_k
3-a_k-1 < 3-a_k
1/(3-a_k-1) > 1/(3-a_k)
and these correlate directly to a_n > a_n+1 for all n.

Does this seem correct?