Answer
Divergent.
Work Step by Step
We have $\mathop \sum \limits_{k = 1}^\infty \sin \dfrac{{k\pi }}{2}$.
Let ${a_k} = \sin \dfrac{{k\pi }}{2}$. We examine the series using the Divergence Test (Theorem 9.4.1):
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \sin \dfrac{{k\pi }}{2} = \left\{ {\begin{array}{*{20}{c}}
{\dfrac{1}{2},}&{k = 1,5,9,\cdot\cdot\cdot}\\
{ - \dfrac{1}{2},}&{k = 3,7,11,\cdot\cdot\cdot}\\
{0,}&{k\ {\rm{ even}}}
\end{array}} \right.$
Since $\mathop {\lim }\limits_{k \to \infty } {a_k} \ne 0$, by the Divergence Test, the series diverges.
