Determine whether the sequence is divergent or convergent.

Question

a(sub n)=(−1)^n(sin(2/n))

Answer 1

|a_n| = |sin(2/n)|

Since |sin(x)| <= |x|, we have:

|a_n| <= 2/n

The limit if |a_n| for n to infinity is thus 0, which then implies that the limit of a_n is zero.

You can easily make this rigorous, for every epsilon > 0, you can using the above inequality find an N such that for every n > N the absolute value of the difference of a_n and the limit is less than epsilon.