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

$\sin \left(\frac{\left(2 n - 1\right) \pi}{2}\right) = \sin \left(n \pi - \frac{\pi}{2}\right) = - \cos \left(n \pi\right) = - {\left(- 1\right)}^{n}$ so
${\sum}_{n = 1}^{\infty} \sin \left(\frac{\left(2 n - 1\right) \pi}{2}\right) = - {\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} / n$
This is an alternating series and ${\lim}_{n \to \infty} \frac{1}{n} = {a}_{n} = 0$ so it is convergent.