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!
 A: $$\sum_{n=1}^\infty \frac{n\sin(\frac{\pi n}{2})}{n^2+12}=\sum_{n=0}^\infty \frac{(-1)^n(2n+1)}{(2n+1)^2+12}$$ The series converges by alternating series test.
A: Determination the sum of the sequence:
$S=\sum\limits_{n=1}^{\infty}\frac{n\sin(\frac{n\pi}{2})}{n^2+12}$
When n is even the value of the sum is equal to zero, othervise: 
$S=\sum\limits_{k=1}^{\infty}\frac{(2k+1)\sin(\frac{(2k+1)\pi}{2})}{(2k+1)^2+12}$
As $\sin(\frac{(2k+1)\pi}{2})=(-1)^k$ and let $x=2\sqrt{3}$
We can form the sum in the followings: 
$S= 4\sum\limits_{k=0}^{\infty}  (-1)^k\frac{2k+1}{(2k+1)^2+x^2}=2  \sum\limits_{k=0}^{\infty}  \frac{(-1)^k}{2k+1-ix}+2\sum\limits_{k=0}^{\infty}\frac{(-1)^k}{2k+1+ix}$
Let $k=2m$ if $k$ even and $k=2m+1$ if $k$ odd so we get: 
$S=2  \sum\limits_{m=0}^{\infty}  \frac{1}{4m+1-ix}-2\sum\limits_{k=0}^{\infty}\frac{1}{4m+3-ix}+\sum\limits_{m=0}^{\infty}  \frac{1}{4m+1+ix}-2\sum\limits_{k=0}^{\infty}\frac{1}{4m+3+ix}$
Introduce the digamma function: $\psi(1+z)=-\gamma+\sum\limits_{k=1}^\infty\big(\frac{1}{k}-\frac{1}{z+k}\big)$
We have the followings: 
$S=\frac{1}{2}\big(-\psi(\frac{ix+1}{4})+\psi(\frac{ix+3}{4})-\psi(\frac{-ix+1}{4})+\psi(\frac{-ix+3}{4})\big)$
Using the reflection formula of digamma function: 
$\psi(1-z)-\psi(z)=\pi \cot(\pi z)$
We get: 
$S=\frac{\pi}{2}\big(\cot(\frac{\pi}{4}(ix+1))+\cot(\frac{\pi}{4}(1-ix))\big)$
After simplifing the expression using trigonometric laws: 
$S=\frac{\pi}{2}\frac{2}{\cos(\frac{ix\pi}{2})}=\pi \sec(ix\frac{\pi}{2})$
Using that $x=2\sqrt{3}$. 
We get that the sequence is convergent and equal to $\pi \sec(i\pi \sqrt{3}))$ 
