Does the sequence have a convergent subsequence

If xn=(sin(n), cos(n), 1+(-1)^n), does the sequence {xn} in R^3 have a convergent subsequence?

I think you just have to show that it follows the Bolzano-Weierstrass Theorem but I'm not sure... Help?

Thanks!

Yes, you are correct.

Note that |sin n| ≤ 1, |cos n| ≤ 1, and |1 + (-1)^n| ≤ 1 + 1 = 2.

Therefore, ||x(n)||
= || (sin(n), cos(n), 1+(-1)^n) ||
= √[sin^2(n) + cos^2(n) + (1 + (-1)^n)^2]
= √[1 + (1 + (-1)^n)^2]
≤ √1 + √(1 + (-1)^n)^2, by triangle inequality
= 1 + (1 + (-1)^n), since 1 + (-1)^n ≥ 0
≤ 1 + 2, since 1 + (-1)^n ≤ 1 + 1 = 2
= 3.

Hence, {x(n)} is bounded above by 3 for all n.
==> {x(n)} has a convergent subsequence by Bolzano-Weierstrass Theorem.

I hope this helps!