Determining convergence of a sum

[pk1234]

I'd really appreciate some help with a sum of:

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.

 

[Dick]

I'd really appreciate some help with a sum of:

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.

No, that won't work. Try to think more basic. Suppose |sin n|<1/10. Can you show |sin(n+1)|>1/10? That would mean for every pair of numbers n, n+1 then either |sin n| or |sin n+1|>1/10. What could you do with that?

 

[pk1234]

Thanks.

If I prove that, should I then take a series of:
b_k = a_2k-1 + a_2k, and say that for all k element N, b_k >= (1/10) * (1/2k) = 1/20k, which diverges?

 

[Dick]

Thanks.

If I prove that, should I then take a series of:
b_k = a_2k-1 + a_2k, and say that for all k element N, b_k >= (1/10) * (1/2k) = 1/20k, which diverges?

Sure that's the idea. And proving my statement can be done pretty crudely. Just draw a graph of the sine function and it should be easy to convince yourself.