# Finding a convergent subsequence does the sequence need to be bounded

• ppy
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

## Homework Statement

2.11. Determine (explicitly) a convergent subsequence of the sequence in R2 given for n =
1; 2; : : : by
xn =(e$^{n}$sin(n$\pi$/7),((4n+3/3n+4)cos(n$\pi$/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$_{n,1}$ not bounded? Therefore surely x$_{n}$ cannot have a convergent subsequence? as doesn't it just go to infinity? Help needed urgently!

Thanks

ppy said:

## Homework Statement

2.11. Determine (explicitly) a convergent subsequence of the sequence in R2 given for n =
1; 2; : : : by
xn =(e$^{n}$sin(n$\pi$/7),((4n+3/3n+4)cos(n$\pi$/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$_{n,1}$ not bounded? Therefore surely x$_{n}$ 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.

1 person
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$_{n,1}$ not bounded? Therefore surely x$_{n}$ 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.