Show that $\displaystyle \sum_{k \ge 0}\frac{1}{k^2+k-\alpha} = \frac{\pi}{\sqrt{4\alpha+1}}\tan\left(\frac{1}{2}\pi\sqrt{4\alpha+1}\right)$ For $\alpha\in\mathbb{R}$, it seems that we have $\displaystyle \sum_{k \ge 0}\frac{1}{k^2+k-\alpha} = \frac{\pi}{\sqrt{4\alpha+1}}\tan\left(\frac{1}{2}\pi\sqrt{4\alpha+1}\right).$ 
I've tried many techniques and ways -- partial fractions, turning it into an integral, finding the partial sum and so on -- to show it, but all ended in failure. Many thanks for the help in advance. 

First we note that
$$ S = \sum_{k \ge 0} \frac{1}{k^2+k-\alpha} =
\sum_{k \ge 0} \frac{1}{(k+1/2)^2 - (4\alpha+1)/4} =
4\sum_{k \ge 0} \frac{1}{(2k+1)^2 -a^2},$$
where $a=\sqrt{4\alpha+1},$ and so
$$ S = 4 \sum_{m \textrm{ odd}} \frac{1}{m^2 - a^2}.$$
Now we use the well-known cotangent identity (at the bottom of the page)
$$\pi\cot(a\pi) = \frac{1}{a} + \sum_{m=1}^\infty \frac{2a}{a^2-m^2} . \quad (1)$$
Replacing $a$ by $a/2$ and dividing by $2$ gives
$$ \frac12 \pi\cot \left( \frac{a\pi}{2}\right) = \frac{1}{a} + \sum_{m=1}^\infty \frac{2a}{a^2-(2m)^2} . \quad (2)$$
Subtracting $(2)$ from $(1)$ gives 
$$ \pi \cot(a\pi) - \frac12 \pi\cot \left( \frac{a\pi}{2}\right) =
\sum_{m=1, \textrm{ odd } m}^\infty \frac{2a}{a^2-m^2} $$
but $\cot(\theta) -\frac12 \cot(\theta/2)= -\frac12 \tan(\theta/2)$ and so
$$ \frac{\pi}{4a} \tan \left( \frac{a \pi}{2} \right) = 
\sum_{m \textrm{ odd}} \frac{1}{m^2 - a^2},$$
from which the result follows setting $a = \sqrt{4\alpha + 1}.$
EDIT: The cotangent identity is proven here.
EDIT2: An easy way to discover the cotangent identity is to take logarithms
of the following product formula for $\sin  \pi x$ and differentiate wrt $x.$
$$\sin \pi x = \pi x \prod_{n=1}^{\infty} \left( 1 - \frac{x^2}{n^2} \right)$$
