- #1

pk1234

- 11

- 0

a_n= |sin n| / n

All I've thought of, is that I should probably create a subsequence of {a_n}, such that all the elements of this subsequence {a_n_k} are >epsilon >0, and then compare the subsequence to 1/n which diverges.

However, I have no idea how to go about this. I don't know how to show, that there really is 'enough' of >epsilon>0 terms. I can't think of a pattern by which to choose only those n, for which a_n > epsilon>0, and I don't really know whether that's possible. If I knew that | sin n| > 0.1 for all n=4k -2 or something like that, it would probably be easy, but since that seems impossible, I don't know what to do.