$\displaystyle \sin\pi/2 = 1,\ \sin2\pi/2=0,\ \sin3\pi/2=-1,\ \sin4\pi/2=0,\ \sin5\pi/2=-1$, and in general $\displaystyle \sin(2m\pi/2)=0,\ \sin((2m+1)\pi/2) = (-1)^m.$
Therefore in the sum $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}}n\sin(n\pi/2)$, only the odd-numbered terms are nonzero. So try putting n=2m+1, with the sum going from m=1 to ∞.