tough series involving digamma I ran across a series that is rather challenging.  For kicks I ran it through Maple and it gave me a conglomeration involving digamma. Mathematica gave a solution in terms of Lerch Transcendent, which was worse yet.  Perhaps residues would be a better method?.
But, it is $$\sum_{k=1}^{\infty}\frac{(-1)^{k}(k+1)}{(2k+1)^{2}-a^{2}}.$$
The answer Maple spit out was:
$$\frac{a+1}{16a}\left[\psi\left(\frac{3}{4}-\frac{a}{4}\right)-\psi\left(\frac{-a}{4}+\frac{1}{4}\right)\right]+\frac{a-1}{16a}\left[\psi\left(\frac{3}{4}+\frac{a}{4}\right)-\psi\left(\frac{1}{4}+\frac{a}{4}\right)\right]+\frac{1}{a^{2}-1}.$$
Is it possible to actually get to something like this by using $\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{k+a}\right]=\gamma+\psi(a+1)?$
I tried, but to no avail.  But, then again, maybe it is too cumbersome. 
i.e. I tried expanding it into 
$\frac{k+1}{(2k+1)^{2}-a^{2}}=\frac{-1}{4(a-2k-1)(2k+1)}-\frac{1}{4(a-2k-1)}+\frac{1}{4(a+2k+1)(2k+1)}+\frac{1}{4(a+2k+1)}$
then using $\sum_{k=1}^{\infty}\left[\frac{1}{k}-\frac{1}{k-\frac{1}{4}-\frac{a}{4}}\right]=\psi\left(\frac{3}{4}-\frac{a}{4}\right)$ and so on, but it did not appear to be anywhere close to the given series. 
On another point, can it be done using residues?. By using $$\frac{\pi csc(\pi z)(z+1)}{(2z+1)^{2}-a^{2}}.$$
This gave me a residue at $\frac{a-1}{2} and  \frac{-(a+1)}{2}$ of 
$\frac{-\pi}{a-1}sec(a\pi/2)$ and $\frac{\pi}{a+1}sec(\pi a/2)$
Taking the negative sum of the residues, it is $\frac{2\pi}{(a-1)(a+1)}sec(a\pi/2)$
By subbing in k=0 into the series, it gives $\frac{-1}{a^{2}-1}$.
I try adding them up and finding the sum, but it does not appear to work out.
Any suggestions?.  Perhaps there is another method I am not trying?.  There probably is. Thanks a million. 

First, group the consecutive oscillating terms together:
$$\sum_{k=1}^{\infty}\frac{(-1)^{k}(k+1)}{(2k+1)^{2}-a^{2}}=\sum_{k=0}^\infty \left(\frac{2k+1}{(4k+1)^2-a^2}-\frac{2k+2}{(4k+3)^2-a^2}\right)-\frac{2(0)+1}{(2(0)+1)^2-a^2}$$
Next, invoke partial fraction decomposition and solve for coefficients:
$$
\frac{2k+1}{(4k+1)^2-a^2}  = \frac{a+1}{16a}\frac{1}{k+\frac{1-a}{4}}+\frac{a-1}{16a}\frac{1}{k+\frac{1+a}{4}},$$
and similarly
$$\frac{2k+2}{(4k+3)^2-a^2}=\frac{a+1}{16a}\frac{1}{k+\frac{3-a}{4}}+\frac{a-1}{16a}\frac{1}{k+\frac{3+a}{4}}.$$
Hence we are left with
$$\frac{a+1}{16a}\sum_{k=0}^\infty \left(\frac{1}{k+\frac{1-a}{4}}-\frac{1}{k+\frac{3-a}{4}}\right)+\frac{a-1}{16a}\sum_{k=0}^\infty\left(\frac{1}{k+\frac{1+a}{4}}-\frac{1}{k+\frac{3+a}{4}}\right)+\frac{1}{a^2-1}$$
$$=\begin{array}{c} \frac{a+1}{16a}\sum_{k=0}^\infty \left(\left(\frac{1}{k+1}-\frac{1}{k+\frac{3-a}{4}}\right)-\left(\frac{1}{k+1}-\frac{1}{k+\frac{1-a}{4}}\right)\right) \\ +\frac{a-1}{16a}\sum_{k=0}^\infty\left(\left(\frac{1}{k+1}-\frac{1}{k+\frac{3+a}{4}}\right)-\left(\frac{1}{k+1}-\frac{1}{k+\frac{1+a}{4}}\right)\right)+\frac{1}{a^2-1} \end{array}$$
$$=\frac{a+1}{16a}\left[\psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{1-a}{4}\right)\right]+\frac{a-1}{16a}\left[\psi\left(\frac{3+a}{4}\right)-\psi\left(\frac{1+a}{4}\right)\right]+\frac{1}{a^{2}-1}.$$
