TITLE: tough series involving digamma
QUESTION [9 upvotes]: 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.

REPLY [7 votes]: Note that
$$
\frac{1}{2k+1-a}+\frac{1}{2k+1+a}=\frac{4k+2}{(2k+1)^2-a^2}\tag{1}
$$
and
$$
\frac1a\left(\frac{1}{2k+1-a}-\frac{1}{2k+1+a}\right)=\frac{2}{(2k+1)^2-a^2}\tag{2}
$$
Adding $(1)$ and $(2)$ and dividing by $4$ yields
$$
\begin{align}
\frac{k+1}{(2k+1)^2-a^2}
&=\frac{1+a}{8a}\frac{1}{k+(1-a)/2}-\frac{1-a}{8a}\frac{1}{k+(1+a)/2}\\
&=\hphantom{+ }\frac{1-a}{8a}\left(\frac1k-\frac{1}{k+(1+a)/2}\right)\\
&\hphantom{= }-\frac{1+a}{8a}\left(\frac1k-\frac{1}{k+(1-a)/2}\right)\\
&\hphantom{= }+\frac{1}{4k}\tag{3}
\end{align}
$$

Now, using
$$
\psi(a+1)+\gamma=\sum_{k=1}^\infty\frac{1}{k}-\frac{1}{k+a}\tag{4}
$$
we get
$$
\frac12\psi\left(\frac{a}{2}+1\right)+\frac\gamma2=\sum_{k=1}^\infty\frac{1}{2k}-\frac{1}{2k+a}\tag{5}
$$
and subtracting twice $(5)$ from $(4)$ gives
$$
\psi(a+1)-\psi\left(\frac{a}{2}+1\right)=\sum_{k=1}^\infty(-1)^{k-1}\left(\frac{1}{k}-\frac{1}{k+a}\right)\tag{6}
$$
Furthermore,
$$
\log(2)=\sum_{k=1}^\infty(-1)^{k-1}\frac1k\tag{7}
$$

Using $(3)$, $(6)$, and $(7)$, we get
$$
\begin{align}
\sum_{k=1}^\infty(-1)^k\frac{k+1}{(2k+1)^2-a^2}
&=-\frac{1-a}{8a}\left(\psi\left(\frac{3+a}{2}\right)-\psi\left(\frac{5+a}{4}\right)\right)\\
&\hphantom{= }+\frac{1+a}{8a}\left(\psi\left(\frac{3-a}{2}\right)-\psi\left(\frac{5-a}{4}\right)\right)\\
&\hphantom{= }-\frac14\log(2)\tag{8}
\end{align}
$$
Equivalence of Forms:
Using $(4)$, $(5)$, and $(7)$, we get
$$
\begin{align}
\sum_{k=1}^\infty\frac{1}{2k-1}-\frac{1}{2k-1+a}
&=\color{green}{\sum_{k=1}^\infty\frac{1}{2k-1}-\frac{1}{2k}}+\color{red}{\sum_{k=1}^\infty\frac{1}{2k}-\frac{1}{2k-1+a}}\\
&=\color{green}{\log(2)}+\color{red}{\frac12\psi\left(\frac{a+1}{2}\right)+\frac\gamma2}\tag{9}
\end{align}
$$
Adding $(5)$ to $(9)$ yields
$$
\begin{align}
\psi(a+1)+\gamma
&=\hphantom{+}\log(2)+\frac12\psi\left(\frac{a+1}{2}\right)+\frac\gamma2\\
&\hphantom{= }+\frac12\psi\left(\frac{a}{2}+1\right)+\frac\gamma2\tag{10}
\end{align}
$$
Rearranging $(10)$ shows that
$$
\psi(a)=\log(2)+\frac12\psi\left(\frac{a}{2}\right)+\frac12\psi\left(\frac{a+1}{2}\right)\tag{11}
$$
Applying $(11)$ gives
$$
\psi\left(\frac{3+a}{2}\right)=\log(2)+\frac12\psi\left(\frac{3+a}{4}\right)+\frac12\psi\left(\frac{5+a}{4}\right)\tag{12}
$$
and
$$
\psi\left(\frac{3-a}{2}\right)=\log(2)+\frac12\psi\left(\frac{3-a}{4}\right)+\frac12\psi\left(\frac{5-a}{4}\right)\tag{13}
$$
Plug $(12)$ and $(13)$ into $(8)$
$$
\begin{align}
\sum_{k=1}^\infty(-1)^k\frac{k+1}{(2k+1)^2-a^2}
&=\hphantom{+}\frac{a-1}{16a}\left(\psi\left(\frac{3+a}{4}\right)-\psi\left(\frac{5+a}{4}\right)\right)\\
&\hphantom{= }+\frac{a+1}{16a}\left(\psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{5-a}{4}\right)\right)\\
&=\hphantom{+}\frac{a-1}{16a}\left(\psi\left(\frac{3+a}{4}\right)-\psi\left(\frac{1+a}{4}\right)-\frac{4}{1+a}\right)\\
&\hphantom{= }+\frac{a+1}{16a}\left(\psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{1-a}{4}\right)-\frac{4}{1-a}\right)\\
&=\hphantom{+}\color{red}{\frac{a-1}{16a}\left(\psi\left(\frac{3+a}{4}\right)-\psi\left(\frac{1+a}{4}\right)\right)}\\
&\hphantom{= }\color{red}{+\frac{a+1}{16a}\left(\psi\left(\frac{3-a}{4}\right)-\psi\left(\frac{1-a}{4}\right)\right)}\\
&\hphantom{= }\color{red}{+\frac{1}{a^2-1}}\tag{14}
\end{align}
$$