Find the $a$ to make the sum of series equal to zero,$\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$

Question

Find all values of $a$ which make the sum of series $\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$

Answer 1

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.

Answer 2

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

$$\frac1{3}\sum_{n=1}^{\infty }\frac{1}{n^2+n-3/4}=\frac1{6}\sum_{n=1}^{\infty }\left[\frac{1}{n-1/2}-\frac{1}{n+1/2}\right]+\frac1{6}\sum_{n=1}^{\infty }\left[\frac{1}{n+1/2}-\frac{1}{n+3/2}\right]\\=\frac{1}{6}(2+2/3)=\frac{4}{9}$$