Evaluation of a digamma series involving golden-ratio Let $\varphi =\frac{1}{2} \left(\sqrt{5}+1\right), a=\tan \left(\frac{\sqrt{5} \pi }{2}\right)$, then how can one prove
$$\sum _{n=1}^{\infty } \frac{\psi ^{(0)}(n+\varphi)-\psi ^{(0)}\left(n-\frac{1}{\varphi}\right)}{n^2+n-1}=\frac{\pi ^2 a^2}{\sqrt{5}}+\frac{4 \pi  a}{5}+\frac{\pi ^2}{2 \sqrt{5}}$$
Note that $n^2+n-1=(n+\varphi) \left(n-\frac{1}{\varphi}\right)$. Maybe we should consider the generalized sum i.e. $\sum _{n=1}^{\infty } \frac{\psi ^{(0)}(n+t)-\psi ^{(0)}(n+s)}{(n+s) (n+t)}$? Any help will be appreciated.
 A: We can use the representation
\begin{equation}
\psi(a)-\psi(b)=(a-b)\sum_{p=0}^\infty\frac{1}{(p+a)(p+b)}
\end{equation} 
to express the series
\begin{align}
S(s,t)&=\sum _{n=1}^{\infty } \frac{\psi (n+t)-\psi (n+s)}{(n+s) (n+t)}\\
&=(t-s)\sum _{n=1}^{\infty }\sum_{p=0}^\infty\frac{1}{(p+n+t)(p+n+s)} \frac1{(n+s) (n+t)}\\
&=(t-s)\sum _{n=1}^{\infty }\sum_{k=n}^\infty\frac{1}{(k+t)(k+s)} \frac1{(n+s) (n+t)}
\end{align}
The case $t=\phi, s=-1/\phi$ is explicitely solved in this question.
For the general case, as shown in the answers, and with $f(n)=1/\left((n+s) (n+t)  \right)$, we have
\begin{equation}
\left[\sum_{n=1}^\infty f(n)\right]^2=2\sum_{n=1}^\infty \sum_{k=n}^{\infty} f(n)f(k) - \sum_{n=1}^{\infty}f(n)^2
\end{equation} 
then,
\begin{equation}
S(s,t)=\frac{t-s}{2}\left[\sum _{n=1}^{\infty }\frac1{(n+s) (n+t)}\right]^2+\frac{t-s}{2}\sum _{n=1}^{\infty }\frac1{(n+s)^2 (n+t)^2}
\end{equation} 
With
\begin{equation}
\frac1{(n+s)^2 (n+t)^2}=\frac{1}{(t-s)^2}\left[\frac{1}{(n+s)^2}+\frac{1}{(n+t)^2}\right]+\frac{2}{(t-s)^3}\left[\frac{1}{n+t}-\frac{1}{n+s}\right]
\end{equation} 
and using again the representations in terms of the polygamma functions given above, we obtain
\begin{align}
S(s,t)=\frac{1}{2(t-s)}&\left[\psi(t+1)-\psi(s+1)\right]^2\\
&+\frac{1}{2(t-s)}\left[\psi^{(1)}(t+1)+\psi^{(1)}(s+1)\right]\\
&-\frac{1}{(t-s)^2}\left[\psi(t+1)-\psi(s+1)\right]
\end{align} 
A: Incomplete answer:

Since we have
$$\frac1\varphi=\varphi-1$$
It follows that
$$\psi(n+\varphi)-\psi\left(n-\frac1\varphi\right)=\psi(n+\varphi)-\psi(n-\varphi+1)=-\pi\cot(\pi\varphi)+\sum_{k=-n}^{n-1}\frac1{k+\varphi}$$
By recurrence and reflection formulas. We also have
$$\sum_{n=1}^\infty\frac1{n^2+n-1}=1+\frac\pi2\tan\left(\frac{\sqrt5}2\pi\right)$$
so it remains to tackle
$$\sum_{n=1}^\infty\frac1{n^2+n-1}\sum_{k=-n}^{n-1}\frac1{k+\varphi}$$
