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

## 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 Helper

## 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
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##)?