Does the Limit of sin(n*alpha) as n Approaches Infinity Diverge?

[AndrejN96]

Homework Statement

Show that lim n->inf sin(n*alpha), 0 < alpha < pi, diverges.

Homework Equations

lim n-> inf sin(n) diverges

The Attempt at a Solution

I know how to solve this for a constant value of alpha (i.e pi/3), but am unaware of solving this one, where alpha may vary. I know that sin(alpha) when alpha is in the given interval has a positive value, but don't know how to apply that to this solution.

 

[HallsofIvy]

If you can do it for specific values of alpha, you should be able to use the same method for any value of alpha (you say "am unaware of solving this one, where alpha may vary"- alpha is NOT varying in a specific sequence- it can be any number but is a fixed number).

By the way- you want to show that the sequence diverges, that it has NO limit, not that "the limit diverges".

 

[Ray Vickson]

Homework Statement

Show that lim n->inf sin(n*alpha), 0 < alpha < pi, diverges.

Homework Equations

lim n-> inf sin(n) diverges

The Attempt at a Solution

I know how to solve this for a constant value of alpha (i.e pi/3), but am unaware of solving this one, where alpha may vary. I know that sin(alpha) when alpha is in the given interval has a positive value, but don't know how to apply that to this solution.

As HallsofIvy has indicated, limits do not "converge" or "diverge"; they just either exist or do not exist. However, sequences can converge or diverge.