Finding a convergent subsequence does the sequence need to be bounded

In summary, the problem is asking to determine a convergent subsequence for the sequence xn = (e^n sin(nπ/7), (4n+3)/(3n+4)cos(nπ/3)) in R2 for n = 1, 2, ... The Bolzano-Weierstrass theorem states that every bounded sequence has a convergent subsequence, but checking each coordinate individually shows that x_n is not bounded. However, this does not necessarily mean that x_n cannot have a convergent subsequence. By looking at the first component, which is unbounded due to e^n, we can see that the only possibility for a convergent subsequence is when sin(nπ/7) pulls the
  • #1

ppy

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Homework Statement



2.11. Determine (explicitly) a convergent subsequence of the sequence in R2 given for n =
1; 2; : : : by
xn =(e[itex]^{n}[/itex]sin(n[itex]\pi[/itex]/7),((4n+3/3n+4)cos(n[itex]\pi[/itex]/3))



I know that the Bolzano-weierstrass theorem says that every bounded sequence has a convergent subsequence. I thought I should first check that the xn is bounded by checking each individual co-ordinate. however isn't x[itex]_{n,1}[/itex] not bounded? Therefore surely x[itex]_{n}[/itex] cannot have a convergent subsequence? as doesn't it just go to infinity? Help needed urgently!

Thanks
 
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  • #2
ppy said:

Homework Statement



2.11. Determine (explicitly) a convergent subsequence of the sequence in R2 given for n =
1; 2; : : : by
xn =(e[itex]^{n}[/itex]sin(n[itex]\pi[/itex]/7),((4n+3/3n+4)cos(n[itex]\pi[/itex]/3))



I know that the Bolzano-weierstrass theorem says that every bounded sequence has a convergent subsequence. I thought I should first check that the xn is bounded by checking each individual co-ordinate. however isn't x[itex]_{n,1}[/itex] not bounded? Therefore surely x[itex]_{n}[/itex] cannot have a convergent subsequence? as doesn't it just go to infinity? Help needed urgently!

Thanks

Being unbounded doesn't necessarily mean a sequence has no convergent sequences. Think about the x and y coordinates separately. Can you find a convergent sequence of the x coordinate.
 
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  • #3
ppy said:
I know that the Bolzano-weierstrass theorem says that every bounded sequence has a convergent subsequence. I thought I should first check that the xn is bounded by checking each individual co-ordinate. however isn't x[itex]_{n,1}[/itex] not bounded? Therefore surely x[itex]_{n}[/itex] cannot have a convergent subsequence?

The B-W theorem doesn't say that if a sequence has a convergent subsequence it is bounded. Look at the first component. We know a priori that ##e^n## is headed for ##\infty##. So the only hope for a convergent subsequence on that component is to pick values of n where the sin pulls things down. What do you know about sin(n##\pi##)?
 

1. Does a sequence need to be bounded in order to have a convergent subsequence?

Yes, a sequence must be bounded in order to have a convergent subsequence. This means that the values in the sequence cannot grow infinitely large or infinitely small.

2. What is a convergent subsequence?

A convergent subsequence is a subset of a sequence that approaches a specific limit or value. This means that as you take more terms from the original sequence, the terms in the subsequence get closer and closer to the limit.

3. Can a sequence have multiple convergent subsequences?

Yes, a sequence can have multiple convergent subsequences. This can happen if the sequence has more than one limit or if the sequence is oscillating between two values.

4. Does every bounded sequence have a convergent subsequence?

No, not every bounded sequence has a convergent subsequence. A sequence may be bounded but still not have a specific limit or value that it approaches. In this case, the sequence does not have a convergent subsequence.

5. Is a convergent subsequence always part of the original sequence?

No, a convergent subsequence does not have to be part of the original sequence. It is a subset of the original sequence, but it may skip some terms or rearrange them in a different order. As long as the terms in the subsequence approach a limit, it is considered a convergent subsequence.

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