Find the $a$ to make the sum of series equal to zero,$\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$ Find all values of $a$ which make the sum of series $\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$
 A: Starting from the representation
$$\pi \cot \pi z = \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{z-n},$$
we obtain
\begin{align}
\pi \tan \pi z &= -\pi \cot \pi\left(z-\tfrac{1}{2}\right)\\
&= \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{n+\frac{1}{2}-z}\\
&= \lim_{N\to\infty} \left(\frac{1}{N+\frac{1}{2}-z} + \sum_{n=0}^{N-1} \frac{2z}{\left(n+\frac{1}{2}\right)^2-z^2}\right)\\
&= 2z \sum_{n=0}^\infty \frac{1}{\left(n+\frac{1}{2}\right)^2-z^2}
\end{align}
and
$$\frac{\pi \tan \pi z}{6z} = \sum_{n=0}^\infty \frac{1}{3n^2+3n - 3\left(z^2-\frac{1}{4}\right)}.$$
It should not be hard to find the values of $a$ from that.
A: Here is one solution: $a = 9/4$.
Then
$$\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=-4/9 + \frac1{3}\sum_{n=1}^{\infty }\frac{1}{n^2+n-3/4}=0$$
Note that 
$$\frac1{3}\sum_{n=1}^{\infty }\frac{1}{n^2+n-3/4}=\frac1{6}\sum_{n=1}^{\infty }\left[\frac{1}{n-1/2}-\frac{1}{n+1/2}\right]+\frac1{6}\sum_{n=1}^{\infty }\left[\frac{1}{n+1/2}-\frac{1}{n+3/2}\right]\\=\frac{1}{6}(2+2/3)=\frac{4}{9}$$
