Finding a convergent subsequence does the sequence need to be bounded

[ppy]

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

 

[Dick]

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.

 

[brmath]

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$\pi$)?