Question

How do you determine the convergence or divergence of `Sigma 1/nsin(((2n-1)pi)/2)` from `[1,oo)`?

Answer 1

It is convergent.

Explanation:

`sin(((2n-1)pi)/2)=sin(npi-pi/2)=-cos(npi)=-(-1)^n` so

`sum_(n=1)^oosin(((2n-1)pi)/2)=-sum_(n=1)^oo(-1)^n/n`

This is an alternating series and `lim_(n->oo)1/n=a_n=0` so it is convergent.