Divergence or convergence of a sin series

$$\sum_{n=1}^{\infty}\frac{n\sin(\frac{n\pi}{2})}{n^2+12}$$

Calc noob here.
I have this series. From inputting the first few values I realize this is an alternating series. I also knew this as sin is alternating. From seeing the pattern I tried to solve this series by changing the sum to:

$$\sum_{n=1}^{\infty}\frac{(2n-1)(-1)^{n+1}}{8n+5}$$

The first few values are: 1/13 - 3/21 + 5/37 - 7/61 +...

I'm not sure if this is a valid move.

I believe my second sum diverges. As we find from the series divergence test that the limit would not equal 0.

Any help is appreciated!

Answer

$$\sum_{n=1}^\infty \frac{n\sin(\frac{\pi n}{2})}{n^2+12}=\sum_{n=0}^\infty \frac{(-1)^n(2n+1)}{(2n+1)^2+12}$$ The series converges by alternating series test.